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

Let $g(x,y) = \ln f(x,y)$. Note that, $g(x,y) = 2\ln x + 2\ln y - x^2 - 2y^2$ and
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = \dfrac{2}{x} - 2x\\
\dfrac{\partial g}{\partial y} = \dfrac{2}{y} - 4y\
\end{cases}
$$
On the order hand,
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x}\ln f(x,y) = \dfrac{1}{f(x,y)}\dfrac{\partial f}{\partial x}\\
\dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y}\ln f(x,y) = \dfrac{1}{f(x,y)}\dfrac{\partial f}{\partial y}
\end{cases}
$$
Thus, 
$$
\begin{cases}
\dfrac{\partial g}{\partial x} = x^2y^2e^{-x^2 -2y^2}\biggl(\dfrac{2}{x} - 2x\biggr) = 0\\
\dfrac{\partial g}{\partial y} = x^2y^2e^{-x^2 -2y^2}\biggl(\dfrac{2}{y} - 4y\biggr) = 0
\end{cases}
$$
This system, the critical points are $(0,0)$, $(-1,-\frac{1}{\sqrt{2}})$,  $(-1,\frac{1}{\sqrt{2}})$,  $(1,\frac{1}{\sqrt{2}})$,  $(1,-\frac{1}{\sqrt{2}})$.
