Is the function $f$ given by
$f(z) =\left\{ \begin{array}{ll} \frac{(\bar{z})^2}{z}, & \hbox{if }z\neq 0; \\ 0, & \hbox{if }z=0. \end{array} \right.$ differrentiable at $z=0$?
I start by taking the $\lim\limits_{z\to 0} \frac{\frac{(\bar{z})^2}{z}-0}{z-0}=\lim\limits_{z\to 0} \frac{(\bar{z})^2}{z^2}$
Choosing $z=x$, $\bar z=x$, thus $\lim\limits_{x\to 0}\frac{x^2}{x^2}=1$. Also by choosing $z=iy, \bar{z}=-iy$, $\lim\limits_{y\to 0} \frac{-y^2}{-y^2}=1$.
Hence, from the above the function $f$ is differentiable at $z = 0$.