What is the largest/least value of the function $f$ in $\mathbb{R}^2$? What is the largest/least value of the function $f=x^2ye^{-x^2-2y^2}$ in $\mathbb{R}^2$?
We have that $f'_x=y\left(2e^{-x^2-2y^2}x-2e^{-x^2-2y^2}x^3\right)$ and $f'_y=x^2\left(e^{-x^2-2y^2}-4e^{-x^2-2y^2}y^2\right)$. If $x=0$ then $f'_x=f'_y=0$ so $(0,y)$, where $y\in \mathbb{R}^2$ are stationary points.
Further $1-4y^2=0\iff y=+-1/\sqrt{2}$ and $2x-2x^3=0\iff2x(1-x^2)\iff x=0, x=+-1$ so $+-(1,1/\sqrt{2})$ are also stationary points.
The values for the stationary points are $f(0,y)=0$, $f(+-1,1/\sqrt{2})=\frac{e^{-2}}{\sqrt{2}}$ and $f(+-1,-1/\sqrt{2})=-\frac{e^{-2}}{\sqrt{2}}$.
Now how do I prove that $e^{-2}/\sqrt{2}$ is the largest value and $-e^{-2}/\sqrt{2}$  is the smallest? Would be nice with a blueprint that I can use for any function but I guess it doesn't exist.
Edit: Not looking for solutions that splits up the $f$ to $x^2e^{-x^2}ye^{-2y^2}$.
I wrote an answer and hopefully someone can check that it is right.
 A: Let $g(x)=x^2e^{-x^2}$ and $h(y)=ye^{-2y^2}.$ Then $g[\Bbb R]=[0,e^{-1}]$ and $h[\Bbb R]=[-2^{-1/2}e^{-1},2^{-1/2}e^{-1}]$. Therefore $f[\Bbb R^2]=\{uv:u\in g[\Bbb R]\land v\in h[\Bbb R]\}=[-2^{-1/2}e^{-2},2^{-1/2}e^{-2}].$
A: The idea is to make a circle(ball) that contains the critical points. By the extreme value theorem there will be a greatest and least value within that circle. If we can show that the boundary points' f-value approaches 0 as the radius get larger then we have shown that for an arbitrary large circle that contains the critical points they will give greatest and least value. In our case the max and min value are $\frac {e^{-2}}{\sqrt{2}}$ and  $\frac {e^{-2}}{\sqrt{2}}$ respectively.
Let $\begin{cases}x=r\cos\theta \\ y=r\sin\theta \end{cases}$ where $0 \leq\theta\leq 2\pi$. We need to show that $\lim _{r\rightarrow\infty}=r^3\cos^2\theta\sin\theta e^{-r^2\cos^2\theta-2r^2\sin^2\theta}=0$.
Since $-r^2\cos^2\theta -2r^2\sin^2\theta=-r^2(\cos^2\theta+2\sin^2\theta)=-r^2(1+sin^2\theta)\leq -r^2\iff e^{-r^2(1+\sin^2\theta)}\leq e^{-r^2}$ we have that $r^3\cos^2\theta\sin\theta e^{-r^2\cos^2\theta-2r^2\sin^2\theta}\leq r^3e^{-r^2}\rightarrow 0$ as $r\rightarrow \infty$. So we have shown that for large enough r, radius of the circle, the boundary points will have f-values close to 0 and as $r$ gets larger the boundary points of the f-values will get closer to 0. Since we can choose an arbitrarily large $r$ it means that $\frac {e^{-2}}{\sqrt{2}}$ and  $\frac {e^{-2}}{\sqrt{2}}$ are the largest and least value respectively in $\mathbb{R}^2$.
A: The function to be optimized on $\Bbb R^2$ is $f(x,y)=x^2ye^{-x^2-2y^2}$. The system of equations for the critical points $f_x=f_y=0$ is equivalent to the system $2xy(1-x^2)=0$ and $x^2(1-4y^2)=0.$ The solution of the system is $S=\{(0,y)|y\in\Bbb R\}\cup\{A(1,\frac12),B(-1,\frac12), C(1,-\frac12), D(-1,-\frac12)\}$. The values at these critical points: $f(0,y)=0$, $f(A)=f(B)=\frac12e^{-\frac32}$ and $f(C)=f(D)=-\frac12e^{-\frac32}.$ Since, the domain of the function is open and $\lim_{x^2+y^2\rightarrow\infty}f(x,y)=0$ we decide: Global max. of $f(x,y)$ is $\frac12e^{-\frac32}$ at $A,B$ and Global min. of $f(x,y)$ is $-\frac12e^{-\frac32}$ at $C,D.$
Note: WA says no global max and min. Why?
