Critical points of a twodimensional function I have the function $f:\mathbb{R^2}\rightarrow\mathbb{R^2}$, given by $f=(4x^2-y^2)e^{-x^2+y^2}$. What are its critical points?
Edit: I have the points $(0,0),(1,0),(-1,0)$ but how do I know when they are a minimum maximum or saddle point?
Edit 2: are they all saddle points?
My hessian for (0,0) is $\begin{pmatrix}8&0\\0&-2\end{pmatrix}$, for (1,0) is $\begin{pmatrix}-48/e&4/e\\0&6/e\end{pmatrix}$ and for (-1,0) is $\begin{pmatrix}-48/e&-4/e\\0&6/e\end{pmatrix}$
 A: Compute the gradient of the function, namely $\nabla f$ and then consider cases where (identically everywhere) $$\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}  \right)=(0,0)$$
To classify those points, one needs to compute the Hessian of the function. see the following picture representation of critical points.
Classification of Critical Points in Multivariable Calculus.
A: Here is a way:
Let $$D= \begin{bmatrix} 
  f_{xx} & f_{xy} \\
  f_{yx} & f_{yy} \\ 
\end{bmatrix}$$
Now find out the matrix $D$ at the points (critical). If all the eigenvalues come out to be postive, then it is a local minima (similarly local maxima if all are negative). If one of them is negative and one is positive then a saddle
Note:
Most of the calculus books would say that if 
at $(x_0,y_0)$
i) If $f_{xx}f_{yy}-f_{xy}^2 \gt 0$ and if 
a)$$f_{xx} \gt 0$$  then local minima
b)$$f_{xx} \lt 0$$ then local maxima
ii) If $f_{xx}f_{yy}-f_{xy}^2 \lt 0$ then saddle point
iii) If $f_{xx}f_{yy}-f_{xy}^2 = 0$ then test fails
