Let $S=\{z \in \mathbb C : |z|<1\}$. Suppose that $f: S \to \mathbb C$ is differentiable everywhere in its domain. Suppose further than $f(z)$ is never zero.
I want to show that $g(z):=\frac 1 {\overline {f(1/\ \overline z)}}$ is differentiable for $|z|>1$.
My first thought was to apply the composition-of-differentiable-functions-is-differentiable theorem. But, alas, the complex conjugation function is not differentiable.
I also thought of applying a theorem which states that, if $f$ is analytic in some open connected set $D$ which contains a segment of the x-axis and is symmertic about that axis, then $[\forall z \in D,\overline {f(z)}=f(\overline z)] \iff [f(x)$ is real for each $x$ in the segment$]$.
The above theorem would be helpful if $f(x)$ was real for each $x\in (-1,0) \cup (0,1)$, but this isn't necessarily true. For example, it isn't true if $f(z) \equiv iz$.
This is immediate from the fact that if $g$ is holomorphic in a domain $D$ the so is $\overline {f(\overline {z})}$ on it domain $\{\overline {z}:z \in D\}$. [one way to see this is to use power series. Another way is to verify C-R equations].