I'm working through some analysis books, and while working through the section on directional derivatives, I searched here and found this answer, which states
Let $f: R^2 \to R$ be defined by $f(x,y)= \frac{x^3y}{x^4+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. This function is continuous; its directional derivative is defined at each point in every direction; and at each point its directional derivative is a linear function of the direction.
I set about trying to show this at the origin, and here's what I have so far:
We have $g(t) = (x, y) + (v, w)t$, so $g(0) = (x, y)$. To find the directional derivative, I want to find $(f \circ g)'(0) = G'(0)$. Taking the composition and doing the algreba, I get
\begin{align} G(t) &= (f \circ g)(t) \\ &= \frac{(x+vt)^3 (y+wt)}{(x+vt)^4+(y+wt)^2} \\ &= (x+vt)^3 (y+wt) \left[ (x+vt)^4+(y+wt)^2 \right]^{-1} \end{align}
From the Chain Rule wrt $t$: \begin{align} G'(t) &= \frac{3v(x+vt)^2 (y+wt) + w(x+vt)^3}{(x+vt)^4+(y+wt)^2} - \frac{(4v(x+vt)^3 + 2w(y+wt)) (x+vt)^3 (y+wt) } {((x+vt)^4+(y+wt)^2 )^2} \\ \implies G'(0) &= \frac{3vx^2y + wx^3}{x^4+y^2} - \frac{(4vx^3 + 2wy) x^3 y } {(x^4+y^2)^2} \end{align}
Assuming I did all that algebra correctly, I don't see where this gets me. Do I then have to use the definition of a derivative in one dimension and show that $G'(0)$ satisfies differentiability at $0$, i.e. \begin{equation} G(t) = G(0) + G'(0)(t-0) + X(t-0) \end{equation}
and $\lim_{t\to0} \frac{X(t-0)}{t-0} = 0$?
edit: I found this answer that takes another approach, so using that method, where $y$ is the direction vector and $a$ is the point at which we take the derivative, gives me this:
\begin{align} \lim_{h \to 0}\frac{f(\mathbf{a} + h\mathbf{y}) - f(\mathbf{a})}{h} &= \lim_{h \to 0}\frac{f((0,0) + h(y_{1},y_{2})) - f(\mathbf{0})}{h} \\ &= \lim_{h \to 0}\frac{f(hy_{1},hy_{2})}{h} \\ &= \lim_{h \to 0}\frac{(hy_1)^3 (hy_2)}{h((hy_1)^4 + (hy_2)^2)} \\ &= \lim_{h \to 0}\frac{h^4 y_1^3 y_2}{h(h^4y_1^4 + h^2y_2^2)} \\ &= \lim_{h \to 0}\frac{h^4 y_1^3 y_2}{h^3(h^2y_1^4 + y_2^2)} \\ &= \lim_{h \to 0}\frac{h y_1^3 y_2}{(h^2y_1^4 + y_2^2)} \\ &= 0 \end{align}
Since this limit exists, regardless of the values for the vector $\mathbf{y}$ (assuming that $\mathbf{y}$ isn't the zero vector, can I conclude that the directional derivative exists for all nonzero vectors?
