Locate and classify critical points of a function?

Question

Assuming $f(x,y) = 24xy + 4xy^2 + 3x^3$ and taking partial differentiation respect to $x$ and $y$ I found to be:

$$\dfrac{\partial f}{\partial x} = 24y + 4y^2 +9x^2 \\ \dfrac{\partial f}{\partial y} = 24x + 8xy$$

Answer 1

You have $$\begin{align} \dfrac{\partial f}{\partial x} &= 24y + 4y^2 +9x^2 = 0 \\ \dfrac{\partial f}{\partial y} &= 24x + 8xy = 0. \end{align} $$ You see that $(0,0)$ is a critical point. Also, if $y = 0$ then $x=0$. If $y=0$ then $x = 0$ or $24 + 4y = 0 \Rightarrow y = -6$. So $(0,-6)$ is a critical point.

Assume that $x\neq 0$ (you can handle those cases alone). So from the second equation you get $$ y = \frac{24x}{-8x} = -3. $$ Now put that into the first equation and solve for $x$ to see if you get one more critical point.

Answer 2

For sure $x=y=0$ is a critical point. The next one is with $x=0$ and $y=-6$. you have to finde $(x,y)$ such that the partial derivatives are both zero. another case will be as $$24x+8xy=8x(3+y)$$ so $y=-3$