Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of $$ x^3 + 3x^2y - 4y^3 - x + y + 3 = 0 $$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):- $$ 2y + x = ±1 $$ can't be deduced.
Instead another erroneous line $ y = 0 $ is outputted.
Some help will be appreciated.Alternate methods of solution would work too:)
If the slant asymptote exists, we may denote $$k=\lim_{x \to \infty}\dfrac{y}{x}.$$
Then $$0=\lim_{x \to \infty}\left(1+3\cdot \frac{y}{x}-4 \cdot \frac{y^3}{x^3}-\frac{1}{x^2}+\frac{y}{x^3}+\frac{3}{x^3}\right)=1+3k-4k^3=-(k-1)(2k+1)^2.$$
Thus,$$k=1,-\frac{1}{2}.$$
Then $$0=\lim_{x \to \infty} \frac{-4b^3-12b^2x-9bx^2+b+3}{x^2}=-9b.$$
Thus,$$b=0.$$
If $k=-\dfrac{1}{2}$, we denote $y=-\dfrac{x}{2}+b$ and put it into the equation, we have $$-8b^3+12b^2x+2b-3x+6=0.$$
Then $$0=\lim_{x \to \infty} \dfrac{-8b^3+12b^2x+2b-3x+6}{x}=12b^2-3.$$
Thus, $$b=\pm \frac{1}{2}.$$
As a result, the slant asymptotes are $$y=x, y=-\frac{x}{2}\pm \frac{1}{2}.$$
The question does not give details about how to apply the method of How to find asymptotes of implicit function?, to this particular problem. Here is an application of that method:
In homogeneous coordinates $(X:Y:Z)$, where $(x,y)$ corresponds to $(x:y:1),$ the equation is $$ X^3 + 3X^2Y - 4Y^3 - XZ^2 + YZ^2 + 3Z^3 = 0. \tag1 $$
On the line at infinity, $Z = 0,$ yielding $$ X^3 + 3X^2Y - 4Y^3 = 0, $$ which factors to $$ (X - Y) (X + 2 Y)^2 = 0.$$ This is solved by $X = Y$ or $X = -2Y,$ which says that the asymptotes are of the form $x = y + \text{constant}$ or $x = -2y + \text{constant},$ but this does not say what the constants are.
Continuing with the method of How to find asymptotes of implicit function?, the derivative of the left-hand side of Equation $(1)$ is $$ (3X^2 + 6XY - Z^2)dX + (3X^2 - 12Y^2 + Z^2)dY + (2YZ - 2XZ + 9Z^2) dZ.\tag2 $$
Evaluating $(2)$ at $(X:Y:Z) = (1 : 1 : 0)$ produces $ 9 dX - 9 dY + 0dZ, $ from which we eventually find the asymptote $x = y$ (whose constant term is zero).
But evaluating $(2)$ at $(X:Y:Z) = (-2 : 1 : 0)$ gives only $0 dX + 0 dY + 0 dZ,$ from which we cannot derive the equation of a line. (It is not clear how you got $y = 0$.) So we have to find the constants that give the asymptote lines some other way.
Giving
$$ f(x,y) = x^3 + 3 x^2 y - 4 y^3 - x + y - 3 = 0 $$
the asymptotic directions can be explored by substituting $ y = a x + b\;\; $ into $f(x,y)$ giving
$$ f(x,ax+b) = 3 x^2 (a x+b)-4 (a x+b)^3+a x+b+x^3-x-3 = (1+3a-4a^3)x^3+3b(1-4a^2)x^2 +(a(1-12b^2)-1)x+b(1-4b^2)-3 $$
Now the conditions for $f(x,ax+b)$ to have line behavior are
$$ \left\{ \begin{array}{rcl} 3 b-12 a^2 b=0 \\ -4 a^3+3 a+1=0 \\ \end{array} \right. $$
and solving we find two solutions:
$$ \{a =-\frac 12, \forall b\} \cup \{a = 1, b = 0\} $$
so we have
$$ L_1\to y = x\\ L_2\to y = -\frac x2 + b $$
Considering $f(x,y)$ at their asymptotic values it's value should be the same or
$$ \lim_{x\to 0}f(x,-\frac 12 x+b)=\lim_{x\to 0}f(x,x) = -3 $$
so this can be solved by choosing $b$ such that $1-4b^2 = 0\;\;$ giving
$$ L_1\to y = x\\ L_2\to y = -\frac x2 \pm\frac 12 $$
Attached a plot showing in blue $f(x,y) = 0$ and in red $L_1, L_2$
In other answers we find various heuristics leading to the conjecture that there are asymptotes of the form $$y= x+C\ , \qquad y=-{1\over2}x +C\ .$$ Therefore we introduce new coordinates $(u,v)$ in the plane such that their directions become the directions of the axes: $$\left\{\eqalign{ u&=x+2y\cr v&=x-y\cr}\right.\qquad{\rm resp.}\qquad\left\{\eqalign{x&={u+2v\over3}\cr y&={u-v\over3}\cr}\right.\ .\tag{1}$$ The given function then appears as $$g(u,v)=f\left({u+2v\over3},{u-v\over3}\right)=3+(u^2-1)v\ .$$ The equation $g(u,v)=0$ is equivalent with $$v={3\over 1-u^2}\ .$$ In terms of $u$ and $v$ we therefore have the two vertical asymptotes $u=\pm1$ and the horizontal asymptote $v=0$. These can now be expressed again in terms of $x$ and $y$, using $(1)$.
$$x^3 + 3x^2y - 4y^3 - x + y + 3 = 0 \tag 1$$ Another variant of method consists in expending $y(x)$ to series of the form : $$y(x)\simeq ax+b+\frac{c}{x}+\frac{d}{x^2}+... \tag 2$$ Putting it into Eq.$(1)$ transforms it to the form : $$Ax^3+Bx^2+Cx+D+...\simeq 0 \quad\text{which implies}\quad A=B=C=D=…=0$$ where $$A=1+3a-4a^3=0$$ $$B=3b-12ba^2=0$$ Etc. But we don't compute immediately $C$ , $D$ , … in the general case because this will be much simpler latter in each particular case.
The roots of $1+3a-4a^2=0=(1-a)(1+2a)$ are $a=1$ and $a=-\frac12$.
Case $a=1$ :
In this case $B=3b-12ba^2=0$ implies $b=0$, thus $y(x)\simeq x+\frac{c}{x}+\frac{d}{x^2}+...$ Putting it into Eq.$(1)$ leads to : $$C=-9c=0\quad\text{thus}\quad c=0$$ $$D=3-9d=0\quad\text{thus}\quad d=\frac13$$ $$y(x)\simeq x+\frac{1}{3x}+...$$ So, for $x\to\infty\quad \textbf{the straight line} \quad y=x\quad\textbf{is an asymptote.}$
Case $a=-\frac12$ :
In this case $B=3b-12ba^2=0$ is satisfied any $b$, thus $y(x)\simeq -\frac12x+b+\frac{c}{x}+...$ Putting it into Eq.$(1)$ leads to : $$C=6b^2-\frac32 x=0 \quad\text{thus}\quad b=\pm\frac12$$ $$D=3-4b^3+12bc+b=0\quad\text{thus}\quad c=\frac{4b^3-b-3}{12b}$$
Case $a=-\frac12$ and $b=\frac12$ then $c=-\frac12$ : $$y(x)\simeq -\frac12x+\frac12-\frac{1}{2x}+...$$ So, for $x\to\infty\quad \textbf{the straight line} \quad y=-\frac12x+\frac12\quad\textbf{is an asymptote.}$
Case $a=-\frac12$ and $b=-\frac12$ then $c=\frac12$ : $$y(x)\simeq -\frac12x-\frac12+\frac{1}{2x}+...$$ So, for $x\to\infty\quad \textbf{the straight line} \quad y=-\frac12x-\frac12\quad\textbf{is an asymptote.}$
I'd try to give another answer. Recall from wikipedia and the most common definition of asymptotic line, to find an asymptotic line of $f(x)$ as $x\to\infty$, we need to compute
(1) $a=\lim_{x\to\infty}\frac{f(x)}{x}$
(2) $b=\lim_{x\to\infty}(f(x) - ax)$.
Then the asymptotic line would be $y=ax+b$.
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By implicit function theorem when $x$ is large, $y$ can be written as a function of $x$. Let's begin with
$x^3 + 3x^2 y - 4y^3 -x + y + 3 = 0. \qquad(1) $
Then
$1+ 3\frac{y}{x} - 4(\frac{y}{x})^3 - \frac{1}{x^2} + \frac{y}{x}\frac{1}{x^2} + \frac{3}{x^3}=0$
if $\lim_{x\to\infty} \frac{y}{x}=:a$ exists, then it might be a solution of $1+3a-4a^3=0$. Moreover, the little consequence $\lim_{x\to\infty} \frac{y-ax}{x}=0$ is required.
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Then we may try $x^3 + 3x^2(y-ax+ax) -4(y-ax+ax)^3 - x + (y-ax+ax) + 3 = 0$. After simplifying, we obtain
$3(y-ax) - 4(y-ax) (\frac{y-ax}{x})^2 - 12 a(\frac{y-ax}{x})^2 - 12( y-ax)a^2 - \frac{1}{x} + \frac{y-ax}{x}\frac{1}{x} + \frac{a}{x} + \frac{3}{x^2}=0$. Thus
$(y-ax)[3 -4(\frac{y-ax}{x})^2 -12a^2] = 12a(\frac{y-ax}{x})^2 + \frac{1}{x} -\frac{y-ax}{x}\frac{1}{x} - \frac{a}{x} -\frac{3}{x^2}$. Taking limit, we obtain $b(3-12a^2)=0$, where $b=\lim_{x\to\infty} y-ax$, existing.
$$\newcommand{\AFF}{\mathbb{A}}$$ $$\newcommand{\PSP}{\mathbb{P}}$$
Let
$$ f(x,y) = 0 $$
be an affine plane curve in $\AFF^2_k$. How can one find all possible asymptotic lines $a x + b y + c = 0$?
Answer: One embeds $\AFF^2_k$ into $\PSP^2_k$ with the coordinates $(X:Y:Z)$ and computes the intersections of $f(x,y) = 0$ with the infinite line $Z = 0$. Therefore one puts
$$ Z^d f(X/Z, Y/Z) = g(X,Y,Z) = 0 $$
with $d$ chosen such, that $Z$ does not divide the polynomial $g(X,Y,Z)$.
From $g(X,Y,0) = 0$ one gets a list of intersection points $(X_i: Y_i: 0)$. We choose one of it, which we write wlog as $P = (1 : y_0 : 0)$.
So we assume, that $X = 1$ and compute the affine representation of $f(x,y) = 0$ in the patch $X = 1$, that is $g(1, Y/X, Z/X) = 0$ as $p(u, v) = 0$ with $u=Y/X$ und $v = Z/X$.
It is then $u = y_0, v = 0$ the intersection point $P$ from above. We put $u' = u - y_0$ and get $P = (u' = 0, v = 0)$. Therefore
$$ h(u',v) = p(u' + y_0, v) = h_m(u',v) + \cdots + h_d(u',v) $$
with $h_\nu$ homogeneus of degree $\nu$ and $d > 0$. We choose a linear factor $l = a u' + b v = 0$ of $h_d$ and transform back
$$ l = a \,(u - y_0) + b \, v = a \, (Y/X - y_0) + b \, (Z/X). $$
Projectively this is
$$ l = a Y - y_0 X + b Z $$
and so in the affine patch $Z = 1$ the sought after equation for the asymptotic line is
$$ l = a y - y_0 x + b = 0 $$
Example: Let
$$ f(x,y) = x^3 + y^3 - 3 a \, x y = 0 $$
be "Folium Cartesii".
We have
$$ g(X,Y,Z) = X^3 + Y^3 - 3 a X Y Z = 0. $$
From $Z= 0$ follows $P=(1:-1:0)$, therefore $y_0 = -1$.
Further we have $p(u,v) = g(1,u,v) = 1 + u^3 - 3 a u v$ and therefore with $u' = u + 1$ we get for $h$
$$ h(u',v) = p(u' - 1, v) = u'^3 - 3 u'^2 + 3 u' - 3 a (u' - 1) v = u'^3 - 3 u'^2 - 3a u' v + 3 u' + 3 a v. $$
So it is $h_1(u',v) = 3 (u' + a v)$ and $l = u' + a v = u + 1 + a v = Y/X + 1 + a Z/X = 0$, and therefore finally
$$ l = y + x + a = 0. $$
Finally your example, calculated with maple
