Determine the points where $f$ is has a local minimum/maximum. Multivariable calculus question. This is not homework, but it is in my book and I find it hard to solve:

Determine the points where $f$ is has a local minimum/maximum.
  Determine if it strong/weak and absolute/relative and interior
  extremum or boundary extremum.
$$f(x,y)= (x^2-y^2)e^{-x^2-y^2}$$
  The domain of $f$ is $E=\{(x,y): x^2+y^2 \leq 4 \}$

A worked solution would be really appreciated. Or if someone knows some worked solutions of exercises like these (somewhere on the internet), that would be great as well. Or a step-by-step plan how to solve exercises like these. In other words, any help is really appreciated.
 A: Well, on the boundary of $E$, we have $f(x,y)=(2x^2-4)e^{-4}$ (why?). This shouldn't be difficult to maximize/minimize, bearing in mind that $-2\le x\le 2$.
To find critical points on the interior of $E$, start by finding the first partial derivatives of $f$, and setting both to $0$. Solve this system, and see what solutions $(x,y)$ have the property that $x^2+y^2<4$. Once any such critical points have been found, find all the second partials of $f$, and use the second partial derivative test to determine which of them (if any) are extrema.
Classify all the extrema as directed.
A: First you have to use KKT conditions with a slack variable $s\ge 0$
$$max_{x,y}\quad (x^2-y^2)e^{-x^2-y^2}\qquad s.t.\quad x^2+y^2+s= 4$$
and the Lagrangian becomes
$$L=(x^2-y^2)e^{-x^2-y^2}+\lambda\big(x^2+y^2+s- 4\big)$$
The first order conditions are
$$\frac{\partial L}{\partial x}=2xe^{-x^2-y^2}(\lambda e^{x^2+y^2}-x^2+y^2+1)=0$$
$$\frac{\partial L}{\partial y}=2ye^{-x^2-y^2}(\lambda e^{x^2+y^2}-x^2+y^2-1)=0$$
$$\frac{\partial L}{\partial \lambda}=x^2+y^2+s- 4=0$$
$$s\cdot \frac{\partial L}{\partial s}=s\cdot \lambda=0$$
where the last one is KKT condition. 
Let's check the boundary where $s=0$. The solution sets are $(0,\pm2)$ and $(\pm2,0)$
For interior where $\lambda=0$ the solution sets are $(0,\pm1)$ and $(\pm1,0)$
By second derivative test you can find that
$(\pm1,0)$ global max
$(0,\pm1)$ global min
$(\pm2,0)$ local max only on the boundary
$(0,\pm2)$ local min only on the boundary
