Fairly simple question, I want to find this limit
$$\lim_{z \to z_0} \frac{{\overline z}^2-{\overline {z_0}}^2}{z-z_0}$$
The original question was to find the region at which the function $f(z)=\overline z^2$ is differentiable, and if possible find $f'(z)$.
If $z=x+iy$, then from Cauchy-Riemann we get that $f$ is differentiable on the straight line $4ix+4y=0$.
Assume $z_0$ is on that straight line. $f'(z_0)$ is defined as $\lim\limits_{z \to z_0} \dfrac{{\overline z}^2-{\overline z_0}^2}{z-z_0}$ and I'm having problems finding that limit.
Of course that $\lim\limits_{z \to z_0} \dfrac{{\overline z}^2-{\overline z_0}^2}{z-z_0}=\lim\limits_{z \to z_0} \dfrac{({\overline z}-{\overline z_0})(\overline z + \overline z_0)}{z-z_0}$ but that does not seem to help a great deal.
Your final paragraph does help. Since $\bar z+\overline{z_0}\to 2\overline{z_0}$, we need to look at $\lim_{z\to z_0}\frac{\bar z-\overline{z_0}}{z-z_0}$ (provided $z_0\ne 0$). But $\frac{\bar z-\overline{z_0}}{z-z_0}$ is just an arbitrary nonzero number divided by its conjugate, hence an arbitrary element of $S^1$ and the limit does not exist. (You might check what happens if $z=z_0+\epsilon \cdot e^{it}$).