Proving that addition of partial derivatives of square of an holomorphic function equals to four times its first derivative square. I am looking for a hint for the following exercise:        

Assuming $f(z)$ as an entire function, prove that
$$\left(\frac{∂^2}{∂x^2}+\frac{∂^2}{∂y^2}\right)\cdot|f(z)|^2=4\cdot|f'(z)|^2$$

I suppose that solution involves using Cauchy–Riemann equations, but I´m not sure  about how to do it
 A: Let $f(z) = u+iv$ be holomorphic in nature.
Then we have $u_x = v_y$ and $u_y = -v_x$
$$f'(z) = u_x + iv_x$$
$$|f'(z)|^2 = u_x^2+v_x^2$$
$u$ and $v$ being real and imaginary parts of $f(z)$ (holomorphic) satisfy 

$$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = 4\frac{\partial^2}{\partial \bar{z}\partial z}$$

So, 
$$LHS = \bigg(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\bigg)(|f(z)^2|) = 4\frac{\partial^2}{\partial \bar{z}\partial z}(|f(z)^2|) = 4\frac{\partial^2}{\partial \bar{z}\partial z}(f(z)\cdot\overline{f(z)}) $$
$f(z)$ is independent of $\bar{z}$ and $\overline{f(z)}$ is independent of $z$. So, $\overline{f(z)} = \bar{f}(\bar{z})$
$$LHS = 4\frac{\partial}{\partial z}\big[f(z)\big]\cdot\frac{\partial}{\partial \bar{z}}\big[\bar{f}(\bar{z})\big] = 4 f'(z)\cdot\bar{f'}(\bar{z}) = 4\ f'(z)\overline{f(z)} = 4|f'(z)|^2 = RHS$$
A: The left side is, using the product rule,
$$
2(u_x^2+u_y^2)+2u(u_{xx}+u_{yy})+2(v_x^2+v_y^2)+2v(v_{xx}+v_{yy})
$$
now use that $u,v$ are harmonic functions.
The right side is, using the Cauchy-Riemann equations, resp. $f'(z)=\lim_{t\to 0}\frac{f(z+t)-f(z)}{t}=\lim_{t\to 0}\frac{f(z+it)-f(z)}{it}$
$$
4⋅|u_x+iv_x|^2=4⋅|v_y-iu_y|^2=2⋅|u_x+iv_x|^2+2⋅|v_y-iu_y|^2
$$
