$f(z)$ and $\overline{f(\overline{z})}$ simultaneously holomorphic

I know that I am not answering your final question, but anyway... if $g(z)=\overline{f(\overline z)}$, then

$$\begin{align*} g^\prime(a)=&\,\lim_{z\to a}\frac{g(z)-g(a)}{z-a}\\[2mm] =&\,\lim_{z\to a}\frac{\overline{f(\overline z)}-\overline{f(\overline a)}}{z-a}\\[2mm] =&\,\lim_{z\to a}\frac{\overline{f(\overline z)-f(\overline a)}}{\overline{\ \overline{z-a}\ }}\\[2mm] =&\,\lim_{z\to a}\overline{\,\Biggl[\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}\Biggr]}\\[2mm] =&\,\overline{\lim_{z\to a}\,\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}}\\[2mm] =&\,\overline{\lim_{w\to\overline a}\,\frac{f(w)-f(\overline a)}{w-\overline a}}\\[2mm] =&\overline{f^\prime(\overline a)}\,. \end{align*}$$

Thus, $g$ is holomorphic. The converse is proved similarly (or you can use the fact that the transformation $f\mapsto g$ is idempotent, that is, you return to your original function $f$ when applied twice).

Proof based in holomorphic $\implies$ locally power series: $$f(z) = \sum_{n=0}^\infty a_n (z-c)^n,$$ $$ \tilde f(z) = \overline{f(\overline{z})} = \overline{\sum_{n=0}^\infty a_n (\overline{z}-c)^n} = \sum_{n=0}^\infty\overline{a_n (\overline{z}-c)^n} = \sum_{n=0}^\infty\overline{a}_n(z-\overline{c})^n, $$ where the third equality is true because the continuity of conjugation.