Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be holomorphic function and $u:\mathbb{C}\rightarrow \mathbb{R}$ be a $C^\infty$ subharmonic function. Prove that $\Delta(u\circ f)=((\Delta u)\circ f)|f'|^2$.
Suppose $f=g+ih$. Then $f'=g_x+ih_x$. So $|f'|^2=g_x^2+h_x^2$. Moreover, since $f$ is holomorphic, $g,h$ are harmonic. So, $g_{xx}+g_{yy}=0$ and $h_{xx}+h_{yy}=0$.
$\Delta u=u_{xx}+u_{yy}$, So $(\Delta u)\circ f=u_{xx}(f)+u_{yy}(f)$. Hence, $((\Delta u)\circ f)|f'|^2=(u_{xx}(f)+u_{yy}(f))(g_x^2+h_x^2)$.
By chain rule $\Delta(u\circ f)=(u_x\circ f)f_{xx}+(u_{xx}\circ f)f_x^2+(u_y\circ f)f_{yy}+(u_{yy}\circ f)f_y^2$.
Now my questions are:
Is the last expression correct?
How do I expand $f_x$ and $f_{xx}$ by chain rule?
I suppose you are asking how you get rid of the $f_{xx}$ term ? Actually, I think it is better to use real notation and write $f(x,y)=(a(x,y),b(x,y))$. Then $\partial_x a= \partial_y b$ and $\partial_y a=-\partial_x b$ (Cauchy-Riemann).
For example, $\partial_x (u\circ f) = \partial_x [u(a(x,y),b(x,y))] = (\partial_x u) \circ f \; \partial_x a + (\partial_y u)\circ f \; \partial_x b$ and then \begin{align} \partial_{xx}(u\circ f) & = \partial_{xx} [u(a(x,y),b(x,y))] = \\ & (\partial_{xx} u) \circ f \ \ (\partial_x a)^2 + 2(\partial_{xy} u)\circ f \ \ \partial_x a \; \partial_x b + (\partial_{yy} u)\circ f \ \ (\partial_x b)^2 \\ & (\partial_x u) \circ f \ \partial_{xx}a + (\partial_y u) \circ f \ \partial_{xx}b . \end{align}
Do the same for the $\partial_{yy}$ derivative and use Cauchy-Riemann. The mixed derivative vanishes and the rest combines to the claimed result.