Form of a harmonic function Given that $\phi(x^2+y^2)$ is harmonic, where $\phi: (0, \infty)\to \mathbb{R}$, find the form of $\phi$.
I do not know what they mean by form nor could I find anything online... My book says that for the complex function $f(z)= (y^3-3x^2y)+i(-3xy^2+x^3+C)$ the form $f(z) = i(z^3+c)$ is easily verified and is suggested by taking $y=0$.
Thanks for any help, I will keep searching.
EDIT From the help I got below:
Taking $\psi(x,y) = \phi(x^2+y^2)$ then
\begin{align*}
  \psi_x =& \phi_x(x^2+y^2)2x \\
  \psi_xx =& 2\phi_x(x^2+y^2)+4x^2\phi_{xx}(x^2+y^2) \\
  \psi_y =& \phi_y(x^2+y^2)2y \\
  \psi_yy =& 2\phi_y(x^2+y^2) + 4y^2\phi_{yy}(x^2+y^2)
\end{align*}
I can add $\psi_{xx} + \psi_{yy} = 0$ but where does that get me? Where should I go from here?
EDIT2: Using Polar coordinates we know $x^2+y^2 = r^2$ so we have:
\begin{align*}
  \phi_r =& \phi_r(r^2)2r \\
  \phi_{rr} =& 2\phi_r(r^2) + 4r^2\phi_{rr}(r^2) \\
  \phi_\theta =& 0
\end{align*}

Hint: Note that $\phi$ is a function of one variable, hence $\psi_{xx} + \psi_{yy} = 0$ implies
$$4\phi'(x^2+y^2)+4(x^2+y^2)\phi''(x^2+y^2) = 0,$$
i.e.
$$\phi'(t)+t\phi''(t) = 0.$$
