Differential equation for harmonic functions. Show that a harmonic function satisfies the formal differential equation: $$\frac{\partial^2u}{\partial z \partial \overline{z}}=0$$


Question

How is $\frac{\partial u}{\partial z}$ even defined when $u$ isn't constant. It is clear that for any non-constant complex variabled, real valued, analytic function $u$ we have that $\frac{\partial u}{\partial z}$ is not defined. This comes from Cauchy-Riemann equations. So how does this question make sense?
Using the notation that Robert Israel mentioned you can restate the equation as

$$\dfrac{\partial^2u}{\partial z \partial \bar{z}}=\frac{1}{4}\left[\frac{\partial}{\partial x}-i\dfrac{\partial}{\partial y} \right]\left[\frac{\partial u}{\partial x}+i\dfrac{\partial u}{\partial y} \right]=\frac{1}{4}\left[\dfrac{\partial^2u}{\partial x^2}-i\dfrac{\partial^2u}{\partial y\partial x}+i\dfrac{\partial^2u}{\partial x\partial y}+\dfrac{\partial^2u}{\partial y^2}\right].$$

Can you complete it from here?
Using the notation that Robert Israel mentioned you can restate the equation as

$$\dfrac{\partial^2u}{\partial z \partial \bar{z}}=\frac{1}{4}\left[\frac{\partial}{\partial x}-i\dfrac{\partial}{\partial y} \right]\left[\frac{\partial u}{\partial x}+i\dfrac{\partial u}{\partial y} \right]=\frac{1}{4}\left[\dfrac{\partial^2u}{\partial x^2}-i\dfrac{\partial^2u}{\partial y\partial x}+i\dfrac{\partial^2u}{\partial x\partial y}+\dfrac{\partial^2u}{\partial y^2}\right].$$

Can you complete it from here?