Extrema of $f(x,y)=\sqrt{x^2+y^2} \cdot e^{-(x^{2}+y^{2})}$ i have problems solving this task here:
We have a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$,
$$ f(x,y)=\sqrt{x^2+y^2} \cdot e^{-(x^{2}+y^{2})} $$
Calculate the local extrema of $f$.
Decide for all whether it is a strict local minimum or strict local maximum.
Find the global maximum and minimum of $f$.
My main problem is to calculate the local extrema of $f$. Normally i would calculate the partial derivatives and set them 0.
Like:
$\frac{\partial f}{\partial x} = \frac{xe^{-x^{2}-y^{2}}}{\sqrt{x^{2}+y^{2}}}-2xe^{-x^{2}-y^{2}}\sqrt{x^{2}+y^{2}}$
$\frac{\partial f}{\partial x} = \frac{e^{-x^{2}-y^{2}}(-2x^{2}y-2y^{3}+y)}{\sqrt{x^{2}+y^{2}}}$
If you just help me finding the points of the local extrema i would be very happy. Sitting now since a few days on this task.
 A: Hint:
note that the function is symmetric around the $z$ axis, so it can be better studied in cylindrical coordinates.
Using $\sqrt{x^2+y^2}= r$, the function becomes:
$$
z=re^{-r^2}
$$ 
and the  derivative 
$$\frac{\partial z}{\partial r}=e^{-r^2}(1-2r^2)$$
is more simple.  
Can you do from this?
A: Of course you may use partial derivatives, but the algebra gets messy. Notice the symmetry about the $z$ axis. $r^2=x^2+y^2$ suggests we use polar coordinates. Think about $f$ not as a function of our $(x,y)$ coordinates in space but as a function of our distance from the $z$ axis $r$. We have $f: [0,\infty) \to \mathbb{R}$, and:
$f(r)=re^{-r^2}$
It should be easy to find the extreme values now, using single variable calculus.
A: You did right to compute
$$
\frac{\partial f}{\partial x} = \frac{xe^{-x^{2}-y^{2}}}{\sqrt{x^{2}+y^{2}}}-2xe^{-x^{2}-y^{2}}\sqrt{x^{2}+y^{2}}
$$
but then you somehow changed a $x$ to $y$ in the next step. Here it is better you factor $x$ and the exponential term to get
$$
\frac{\partial f}{\partial x} = \frac{xe^{-(x^2+y^2)}(1-2(x^2+y^2))}{\sqrt{x^{2}+y^{2}}}
$$
and
$$
\frac{\partial f}{\partial y} = \frac{ye^{-(x^2+y^2)}(1-2(x^2+y^2))}{\sqrt{x^{2}+y^{2}}}
$$
Since $e^{-(x^2+y^2)}\neq 0$ for all $(x,y)\in\mathbb R^2$ you get
$$
\frac{\partial f}{\partial x}=0\Leftrightarrow x(1-2(x^2+y^2))=0\Leftrightarrow x=0\text{ or }x^2+y^2=\frac12
$$
and
$$
\frac{\partial f}{\partial y}=0\Leftrightarrow y(1-2(x^2+y^2))=0\Leftrightarrow y=0\text{ or }x^2+y^2=\frac12
$$
The set of critical points is

$$\{(0,0)\}\cup\left\{(x,y)\in\mathbb R^2~:~x^2+y^2=\frac12\right\}$$ The set of critical points is here not finite. Maybe that was confusing?

