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?

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.
