Find extremas of $f(x,y) = xy \ln(x^2+y^2), x>0, y>0$ As the title says I need to find extreme values(maximum and minimum) of
$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$
I don't understand how to find critical points of this problem.
I start with finding partial derivative and set derivatives equal to zero. And that is where I am stuck currently. So any help would be appreciated.
So: 
$
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy* ln(x^2+y^2)) = y*ln(x^2+y^2) + \frac{2xy^2}{x^2+y^2}
$
$
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy* ln(x^2+y^2)) = x*ln(x^2+y^2) + \frac{2x^2y}{x^2+y^2}
$
So we have now:
$
\nabla f(x,y) = (0,0) 
$
$
 y*ln(x^2+y^2) + \frac{2xy^2}{(x^2+y^2)^2} = 0
$
And 
$
 x*ln(x^2+y^2) + \frac{2x^2y}{(x^2+y^2)^2} = 0
$
After trying to solve these equations I get that $x=y$ Is that correct?. So I don't understand what are then critical points as It can't be (0,0)?

We look for points for wich $\nabla f=0$
The method you describe is this:
$\dfrac{\partial f}{\partial x}=y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}$
$\dfrac{\partial f}{\partial y}=x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}$
(You need to practice the derivatives :)
Setting the partials to zero we can get the coordinates for those points.
$y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}=0$
$x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}=0$
$\dfrac{y}{x}\ln(x^2+y^2)=\dfrac{x}{y}\ln(x^2+y^2)\implies x^2=y^2\;,x^2+y^2\neq1$
Then, $y=x$ or $y=-x$, dropping the second as $x\gt0$ and $y\gt0$
$x\ln(2x^2)=-\dfrac{2x^3}{2x^2}\implies \ln(2x^2)=-1$ or $x=\pm\dfrac{e^{-1/2}}{\sqrt{2}}$
Dropping the negative solution:
$x=y=\dfrac{e^{-1/2}}{\sqrt{2}}$
(We need to consider the case $x^2+y^2=1$. It's easily seen that the partials doesn't vanish: for those points, $2x^2y=0$, that is not possible with $x\gt0\;,y\gt0$)
