If $f \in C^2$(as a two variable real function) and $\Delta f = \frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}$ then $$\Delta f = 4 \frac{\partial^2f}{\partial z \partial \overline{z}} = 4 \frac{\partial^2f}{\partial \overline{z} \partial z} $$
I tried playing with Cauchy-Riemann equations:
$$ \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} \right) $$
applying $\frac{\partial}{\partial z}$
$$ \frac{\partial^2f}{\partial \overline{z} \partial z} = \frac{1}{4} \left( \frac{\partial^2f}{\partial x^2} +i \frac{\partial^2f}{\partial y \partial x} + i\frac{\partial^2f}{\partial x \partial y} - \frac{\partial^2f}{\partial y^2} \right) $$
But that leads nowhere.
It does not look like a hard task though, how can we prove it?
In the complex case
\begin{align} \frac{\partial}{\partial z} &= \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) \\ \frac{\partial}{\partial\bar{z}} &= \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) \end{align}then you get \begin{equation} \frac{\partial^2}{\partial z \partial\bar{z}} =\frac{1}{4}\Big(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\Big) \end{equation}
I think you are looking for the Wirtinger derivatives ( https://en.wikipedia.org/wiki/Wirtinger_derivatives ).