Let's consider a "new" basis of the partial differential operators (of order 1) on $\mathbb{R^2}\approx\mathbb{C}$ defined by :
$\frac{\partial}{\partial z}:= \frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \overline{z}}:= \frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$
Recall that a function $f$ of class $\mathscr{C^2}$ (in the real sense) is called harmonic if it satisfies $\frac{\partial^2 f}{\partial x^2}(z)+\frac{\partial^2 f}{\partial y^2}(z)=0$.
Let $f$ be a holomorphic function on $U$ (which implies that $f$ is of class $\mathscr{C^2}$ on $U$). Then $\frac{\partial f}{\partial \overline{z}}=0$ implies $\frac{\partial}{\partial z}(\frac{\partial f}{\partial \overline{z}})=0$. Now I want to show that $f$ is harmonic so let's calculate $$\frac{\partial}{\partial z}(\frac{\partial f}{\partial \overline{z}})=\frac{1}{4}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})f=\frac{1}{4}(\frac{\partial^2}{\partial x^2}+\underbrace{i\frac{\partial^2}{\partial x \partial y}-i\frac{\partial^2}{\partial y \partial x}}_\text{A}+\frac{\partial^2}{\partial y^2})f$$ According to my lecture notes $A=0$ because $f\in\mathscr{C^2}(U)$. Obviously $A=0$ implies that f is harmonic but I don't understand why $f\in\mathscr{C^2}(U)$ implies $A=0$.
Why does $C^2$ "obviously" imply harmonic? There are plenty of twice differentiable functions that are not harmonic. Eg: $f(x,y)=x^3$. The argument they are using here is called Clairaut's theorem which says that if a function is twice differentiable, then the second order mixed partial derivatives are equal. Which is exactly the statement that $A$ in your computation is zero.