Expanding the operators \begin{align} \left(r \frac{d}{d r}\right)^2 = \left(r \frac{d}{d r}\right) \left(r \frac{d}{d r }\right) = r \frac{d}{d r} \left(r \frac{d}{d r}\right)\\ \left(\frac{d}{d \theta}\right)^2 = \left(\frac{d}{d \theta}\right) \left(\frac{d}{d \theta}\right) = \frac{d}{d \theta} \left(\frac{d}{d \theta}\right) \end{align} we have \begin{align} \left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) &= r \frac{d}{d r} \left\{r \frac{d \Phi\big(|f(z)|\big)}{d r}\right\} = r \frac{d}{d r}\left\{r\,\Phi'\big(|f(z)|\big) \frac{d}{d r}|f(z)|\right\} \\ \\ &= r \frac{d}{d r}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Re\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\ &= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Re^2\left[\frac{zf'(z)}{f(z)}\right] \\ & \hskip2in + r |f(z)|\,\Phi'\big(|f(z)|\big) \Re'\left[\frac{zf'(z)}{f(z)}\right] \\ \\ \\ \left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) &= \frac{d}{d \theta} \left\{\frac{d \Phi\big(|f(z)|\big)}{d \theta}\right\} = \frac{d}{d \theta}\left\{\Phi'\big(|f(z)|\big) \frac{d}{d \theta}|f(z)|\right\} \\ \\ &= -\frac{d}{d \theta}\left\{|f(z)|\,\Phi'\big(|f(z)|\big) \Im\left[\frac{zf'(z)}{f(z)}\right]\right\}\\ \\ &= |f(z)| \left\{|f(z)|\,\Phi'\big(|f(z)|\big)\right\}' \Im^2\left[\frac{zf'(z)}{f(z)}\right] \\ & \hskip2in - |f(z)|\,\Phi'\big(|f(z)|\big) \Im'\left[\frac{zf'(z)}{f(z)}\right] \end{align} Adding the two \begin{multline} \left(r \frac{d}{d r}\right)^2 \Phi\big(|f(z)|\big) + \left(\frac{d}{d \theta}\right)^2 \Phi\big(|f(z)|\big) = \\ \Psi\big(|f(z)|\big) \left|\frac{zf'(z)}{f(z)}\right|^2 + |f(z)|\,\Phi'\big(|f(z)|\big)\left\{\Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right]\right\} \end{multline}
Finally, using that $f(z)$ is analytic in $D$, you need to prove that the second term is zero. I'm very rusty on my complex analysis theorems, but I believe is pretty straigth forward since $ \Re'\left[\frac{zf'(z)}{f(z)}\right] - \Im'\left[\frac{zf'(z)}{f(z)}\right] = \frac{d}{dz} \left(\frac{\bar{z}\bar{f}'(z)}{\bar{f}(z)}\right) = \frac{d}{dz} \left(\frac{\bar{z} f'(\bar{z})}{f(\bar{z})}\right) = 0 $ Please doublecheck that the last argument is correct.