Form of a harmonic function

Question

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*}

Answer 1

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. $$\phi'(t)+t\phi''(t) = 0.$$

Answer 2

Hint: convert to polar coordinates. The condition for $u(r,\theta)$ to be harmonic is $u_{rr} + u_r/r + u_{\theta,\theta}/r^2 = 0$.

Alternatively, consider the map of $\mathbb C$ onto ${\mathbb C} \backslash \{0\}$ by $w \mapsto z = e^{w}$. $u(z)$ is harmonic on ${\mathbb C} \backslash \{0\}$ if and only if $U(w) = u(e^{w})$ is harmonic on $\mathbb C$. What harmonic functions of $w$ depend only on the real part of $w$?