Considering two regions and a holomorphic function, show the following

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I'm very lost on how to do this question. Do I use cauchy-reimann equations somehow? Thank you for your help!

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We use $\triangle f=4\dfrac{\partial^2f}{\partial z\partial\bar{z}}$.

$$\begin{eqnarray} \triangle (\psi\circ f) &=& 4\dfrac{\partial^2}{\partial z\partial\bar{z}}(\psi\circ f)\\ &=& 4\dfrac{\partial}{\partial z}\left(\dfrac{\partial}{\partial\bar{z}}(\psi\circ f)\right)\\ &=& 4\dfrac{\partial}{\partial z}\left( \dfrac{\partial}{\partial f}(\psi\circ f)\dfrac{\partial f}{\partial\bar{z}}+\dfrac{\partial}{\partial\bar{f}}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \right)\\ &=&4\dfrac{\partial}{\partial z}\left( \dfrac{\partial}{\partial\bar{f}}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \right)\\ &=&4\dfrac{\partial^2}{\partial z \partial\bar f} (\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} + 4\dfrac{\partial}{\partial\bar f} (\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}\partial z} \\ &=&4\dfrac{\partial}{ \partial\bar f}\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial{f}}{\partial{z}} + \dfrac{\partial}{ \partial \bar f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial\bar{f}}{\partial{z}}\right) + 4\dfrac{\partial}{\partial\bar f} (\psi\circ f)\overline{\left(\dfrac{\partial{f}}{\partial\bar{z}\partial z}\right)} \\ &=&4\dfrac{\partial}{ \partial\bar f}\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\dfrac{\partial\bar{f}}{\partial\bar{z}} \dfrac{\partial{f}}{\partial{z}}\right) \\ &=&4\dfrac{\partial}{ \partial\bar f }\left( \dfrac{\partial}{ \partial f}(\psi\circ f)|f'|^2\right) \\ &=&4\dfrac{\partial}{ \partial\bar f }\left( \dfrac{\partial}{ \partial f}(\psi\circ f)\right)|f'|^2 \\ &=&4\dfrac{\partial^2}{ \partial\bar f \partial f }(\psi\circ f))|f'|^2 \\ &=& \triangle \psi .|f'|^2 \end{eqnarray} $$ Lines 4 and 7 we use the fact that $\dfrac{\partial f}{ \partial \bar z}=0$ because $f$ holomorphic