Determing if a function is analytic on a set Let $f$ be analytic on a set $U$.  Let $g(z)=\overline{f(\bar{z})}$.  Is $g(z)$ analytic on $V=\{z\, : \bar{z} \in U\}$.
Is it enough to show this:  $$g'(z)=\lim\limits_{z \to z_0}\frac{g(z)-g(z_0)}{z-z_0}=\lim\limits_{z \to z_0}\frac{\overline{f(\bar{z})}-\overline{f(\bar{z_0})}}{z-z_0}=\lim\limits_{z \to z_0}\overline{\left(\frac{f(\bar{z})-f(\bar{z_0})}{\bar{z}-\bar{z_0}}\right)}$$
and remark that $f$ is analytic on $U$.
 A: The answer is YES!
In particular, the function $g(z)=\overline{f(\bar z)}$ is analytic in $V$, while $f(\bar z)$ is not analytic on $V$.
Why is $g$ analytic on $V$? 
Simply because for every $z_0\in V$ (equivalently $\bar z_0\in U$) the limit
\begin{align}
\lim\limits_{z \to z_0}\frac{g(z)-g(z_0)}{z-z_0} &=\lim\limits_{z \to z_0}\frac{\overline{f(\bar{z})}-\overline{f(\bar{z_0})}}{z-z_0}=\lim\limits_{z \to z_0}\overline{\left(\frac{f(\bar{z})-f(\bar{z_0})}{\bar{z}-\bar{z_0}}\right)}=\lim\limits_{\zeta \to \bar z_0}\overline{\left(\frac{f(\zeta)-f(\bar{z_0})}{\zeta-\bar{z_0}}\right)} \\ &=
\overline{f'(\bar z_0)},
\end{align}
does exist! 
Consequently $\big(\overline{f(\bar z)}\big)'=\overline{f'(\bar z)}$.
Note. This is a standard exercise, and one I always ask whenever I teach C.V. 
A: No. It is not true that $z \mapsto f(\overline{z})$ is analytic on $U$. The expression that you got is the correct way to do it, but now you have to explicitly use differentiability of $f$ at $\overline{z}$, which will give you what you want because the neighbourhood of $\overline{z}$ is the conjugate of the neighbourhood of $z$.
