Local extremes of: $f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$ I am looking to find the local extremes of the following function:
$$f(x,y) = (x^2 + 3y^2)e^{-x^2-y^2}$$
What have I tried so far?


*

*Calculate the partial derivatives:


$$\frac{\partial f}{\partial x} = 2x(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2x)$$
$$=-2(e^{-x^2-y^2})x(-1+x^2+3y^2)$$
$$\frac{\partial f}{\partial y} = 6y(e^{-x^2-y^2}) + (x^2+3y^2)(e^{-x^2-y^2})(-2y)$$
$$=-2(e^{-x^2-y^2})y(-3+x^2+3y^2)$$


*Determine the possible stationary points:
$$\frac{\partial f}{\partial x}=0 \implies -2(e^{-x^2-y^2})x(-1+x^2+3y^2) = 0$$
$$\implies [x=0] \text{ or } -1+x^2+3y^2 = 0 \implies [x=\sqrt{-3y^2+1}]$$


$$\frac{\partial f}{\partial y} = 0 \implies [y=0] \text{ or } -3+x^2+3y^2 = 0 \implies [y=\pm \sqrt{\frac{-x^2+3}{3}}]$$
So now I have:
$$\frac{\partial f}{\partial x}\begin{cases} x = 0 ,\\
\sqrt{-3y^2+1}\end{cases}
\text{ and }
\frac{\partial f}{\partial y}\begin{cases} y = 0 ,\\
\sqrt{\frac{-x^2+3}{3}}\end{cases}$$
From this I determined the following possible points:
$$(0,0) \rightarrow f(0,0) = 0$$
$$(0,1) \rightarrow f(0,1) = 3e^{-1}$$
$$(0,-1) \rightarrow f(0,-1) = -3e^(-1)$$
$$(1,0) \rightarrow f(1,0) = e^{-1}$$
$$(\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
$$(-\sqrt{\frac{1}{3}},\sqrt{3}) \rightarrow f(-\sqrt{\frac{1}{3}},\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
$$(\sqrt{\frac{1}{3}},-\sqrt{3}) \rightarrow f(\sqrt{\frac{1}{3}},-\sqrt{3}) = \frac{28}{3}e^{-\frac{5}{4}}$$
So far, did I follow the steps correctly? Also, from what I know next is to check the diferencial of the function in each point and those which are zeros are extreme points. Is that correct? And what would follow this?
 A: In 1D (single variable Calculus), a stationary-critical point is a point $c$ such that $f'(c)=0$. At $c$ the gradient of $f$ is zero (there is no change). 
In 2D, a stationary-critical point is a point $(a, b)$ such that the gradient in $x$ direction and $y$ direction are both zero (at that same point $(a,b)$, simultaneously, as @Moo said). At that point, no change of value of $f$ in both $x$ and $y$ direction. This means : 
$$ \frac{\partial f(a,b)}{\partial x } = \frac{\partial f(a,b)}{\partial y } = 0  $$
So, in your case, you must find a point $(a, b)$ such that both
$$ \frac{ \partial f }{\partial x} = - 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right)  = 0 $$
$$ \frac{ \partial f }{\partial y} = - 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right)  = 0 $$
which means solving both at once. The system to solve is :


*

*$- 2 e^{-x^{2} - y^{2}}x \left( -1 + x^{2} + 3y^{2} \right)  = 0$

*$- 2 e^{-x^{2} - y^{2}}y \left( -3 + x^{2} + 3y^{2} \right)  = 0$


Since $e^{...}$ is never $0$, you may solve these instead :


*

*$x \left( -1 + x^{2} + 3y^{2} \right)  = 0$

*$y \left( -3 + x^{2} + 3y^{2} \right)  = 0$


You can continue from here to solve the system.
Hope this helps.
