<s>[INST] Global Max and Min Problem I'm working on a problem which asks me to find local and global extrema of the following function.
$$f(x,y) = x^2y^2e^{(-x^2 - 2y^2)}$$
I went through and found all of the relevant partial derivatives.
\begin{align*}
f_x &= (2xy^2)(e^{(-x^2 - 2y^2)}) + (x^2y^2)(e^{(-x^2 - 2y^2)})(-2x)\\
f_x &= (e^{(-x^2-2y^2)})(2xy^2 -2x^3y^2)\\
\\
f_y & = (2x^2y)(e^{(-x^2-2y^2)}) + (x^2y^2)(e^{(-x^2-2y^2)})(-4y)\\
f_y &= (e^{(-x^2-2y^2)})(2x^2y-4x^2y^3)\\
\\
f_{xx} &= (e^{(-x^2-2y^2)})(-2x)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(2y^2 -6x^2y^2)\\
f_{xx} &= (e^{(-x^2-2y^2)})(-10x^2y^2 + 4x^4y^2 + 2y^2)\\
\\
f_{yy} &= (e^{(-x^2-2y^2)})(-4y)(2x^2y-4x^2y^3) + (e^{(-x^2-2y^2)})(2x^2 - 12x^2y^2)\\
f_{yy} &= (e^{(-x^2-2y^2)})(-20x^2y^2 + 16x^2y^4 + 2x^2)\\
\\
f_{xy} &= (e^{(-x^2-2y^2)})(-4y)(2xy^2 -2x^3y^2) + (e^{(-x^2-2y^2)})(4xy-4x^3y)\\
f_{xy} &= (e^{(-x^2-2y^2)})(-8xy^3 + 8x^3y^3 +4xy - 4x^3y)\\
\end{align*}
However, I'm not sure what to do after this. I thought I was  supposed to set $f_x$ and $f_y$ equal to 0 but I don't know how to solve the equations that I get. Can someone please help me? Did I make a mistake while I was determining my partial derivatives?
EDIT: I made a mistake calculating the partial derivatives and I edited that
 [/INST] You have:
$$0= (e^{(-x^2-2y^2)})(2xy^2 -2x^3y^2)$$
$$0=(e^{(-x^2-2y^2)})(2x^2y-4x^2y^3)$$
Because $0 \neq e^{(-x^2-2y^2)}$:
$$0= 2xy^2 -2x^3y^2$$
$$0=2x^2y-4x^2y^3$$
You have two cases:
1)$x \neq 0$, then from second equation $0=y-2y^2$, so $y=0$ or $y=\frac{1}{2}$, so $(z,0)$ for all $z$ is potential extremum, and for $y=\frac{1}{2}$ from first $0=x-x^3$, so $x=1$ or $x=-1$.
2)$x=0$, then $(0,z)$ is potencial extermum for all $z$.
</s>