Knowing: $f(z)$ is analytical
Prove: $$(\frac{\partial^2}{\partial^2 x} + \frac{\partial^2}{\partial^2 y})|f(z)|^2 = 4|f'(z)|^2$$
I have proved firstly that $\ln|f(z)|$ is harmonic function
Let $$f(z) = u(x,y) + i v(x,y)$$
And I transformed what to prove into: $$u(\frac{\partial^2 u}{\partial^2 x} + \frac{\partial^2 u}{\partial^2 y}) + v(\frac{\partial^2 v}{\partial^2 x} + \frac{\partial^2 v}{\partial^2 y}) = 0$$
However, when I was trying to use: $$(\frac{\partial^2}{\partial^2 x} + \frac{\partial^2}{\partial^2 y})\ln|f(z)|=0$$ I came into trouble. This equation is too complex to expand.
So, I supposed that there should be a better idea. It is nice of you to give a hint.
One approach is to rewrite the Laplacian in terms of complex (Wirtinger) derivatives, as in Show that $4\frac{\partial}{\partial z}\frac{\partial}{\partial\bar{z}}=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. The note that in a neighborhood of a point where $|f|\ne 0$ you can write $$\ln |f| = \frac12\ln f+\frac12 \ln \bar f$$ which is the sum of holomorphic and antiholomorphic functions. Now it should be easy to apply the Laplacian in the above form.
Another approach is more real-variable based: for every smooth function $u$ on a domain in $\mathbb R^n$ we have $$\nabla e^{u} = e^u \nabla u$$ hence $$\Delta e^u = \operatorname{div}\nabla e^u = \nabla e^u\cdot \nabla u+e^u \operatorname{div}\nabla u = e^u |\nabla u|^2+e^u\Delta u$$ Apply this to $u=2\ln|f|$ which you know to be harmonic; conclude $$\Delta |f|^2 = 4 |f|^2 |\nabla \ln |f||^2 $$ Then it remains to check that $|\nabla \ln |f||^2 = |f'(z)|^2/|f(z)|^2$.