Find critical points of function $f(x,y)=(x^2+y^2) \ln(x^2+y^2)$ I want to find critical points of $$f(x,y)=\left\{\begin{matrix}
(x^2+y^2) \ln(x^2+y^2) & \text{if} &(x,y)\neq (0,0) \\ 
 0& \text{if}  & (x,y)=(0,0)
\end{matrix}\right.$$
I have started to find first derivatives:
$$f_x=2x(\ln(x^2+y^2)+1),$$
$$f_y=2y(\ln(x^2+y^2)+1).$$
Here it is my problem that I cannot find critical points here. I have tried to do like this:
$$2x(\ln(x^2+y^2)+1)=0,$$ therefore $x=0$ and $\ln(x^2+y^2)=-1$ but this one doesn't seem correct.
 A: When solving problems of optimization, one should always look for transformations that simplify the problem.
if you apply the coordinate trans $$
\begin{cases}
x=\sqrt{r} \sin{t}\\
y=\sqrt{r} \cos{t}\\
\end{cases}
$$
you end up with the function defined in $\mathbf{R}^+$
$$
g(r)=r \log{r}
$$
which has got a minima for $r=\frac{1}{e}$
so the set of minima for the original function is :
$$
M_n=\left\{(\frac{\sin t}{\sqrt e};\frac{\cos t}{\sqrt e})\middle| t \in [0,2\pi]\right\}
$$
which is the circle of radius $\frac{1}{e}$ centered in the origin where the function f attains it's minimum value of $-\frac{1}{e}$
Remark: the origin is a local maximum as can be seen from your partial derivatives
A: Use the polar form of the equation:
$$f(r,\theta)=r^2\ln r^2$$
(where $f$ is zero if $r$ is zero). Since the expression only depends on $r$, $f$ is circularly symmetric about the origin.
We can then differentiate $r^2\ln r^2$ to find critical points; we find two at $r=0$ and $r=e^{-\frac12}$. Hence the critical points of $f$ in the original Cartesian formulation are the origin and the circle with equation $x^2+y^2=\frac1e$. (The origin is a removable singularity.)
