When does $h(x^2+y^2)$ be harmonic?
Solution 1:
$ u=h(x^2+y^2) $ \begin{eqnarray*} \frac{ \partial u} {\partial x} =2x h'(x^2+y^2) \\ \frac{ \partial^2 u} {\partial x^2} = 4x^2 h''(x^2+y^2)+ 2 h'(x^2+y^2). \\ \end{eqnarray*} Substitute this into $u_{xx}+u_{yy}=0$ and let $z=x^2+y^2$ \begin{eqnarray*} \color{blue}{zh''(z)+h'(z)=0}. \end{eqnarray*}
Solution 2:
For every $z>0$, $z=x^2+y^2$ describes a circle with center $(0,0)$. The mean value of $u(x,y)$ over this circle is $h(z)$, and this must be equal to $u(0,0)=h(0)$ due to the mean value property of harmonic functions.