Consider the function defined by
$$f(z)=\begin{cases}0&\text{ if }\operatorname{Re}(z)=0\vee\operatorname{Im}(z)=0\\1&\text{ otherwise.}\end{cases}$$
I would like to show that the real and imaginary parts of $f$ satisfy the Cauchy-Riemann equations at the origin but am not sure if just saying $0=0$ and $0=-0$ is enough or not also is $f$ complex differentiable at the origin, I am not sure on how to show whether it is or not, thank you
Basically, you are right. $\frac{\partial u}{\partial x}(0,0)=\lim_{h\to0}\frac{u(h,0)-u(0,0)}h=\lim_{h\to0}0=0$. By the same kind of argument, $\frac{\partial u}{\partial y}(0,0)=\frac{\partial v}{\partial x}(0,0)=\frac{\partial v}{\partial y}(0,0)=0$. Since $0=0$ and $0=-0$…
Note that $f$ not differentiable at the origin, since it's not even continuous there.