Let $f$ be holomorphic, $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant. Let $f$ be holomorphic, $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.
My proof: Let $$f(\overline{z})=\Sigma a_n(\overline{z}-z_0)^n$$ and we know the series converges uniformly as $|z|=|\overline{z}|$ thus we can take the derivative term wise, but each term for $n>0$ is not differentiable as $\overline{z}$ is not differentiable. Thus the limit does not exist and the function is not differentiable. Unless $f$ is a constant.
Is this correct?
 A: I'm not sure if this line of reasoning will get you there.  It is possible for the sum of two non-differentiable functions to be differentiable. Consider $f(z) = {\rm Re}(z)$ and $g(z) = i{\rm Im}(z)$. Neither is differentiable, but their sum $f(z) + g(z) =z$ is differentiable.
Instead, why not use the Cauchy-Riemann equations? Writing $f(x+iy)= u(x,y)+iv(x,y)$, what do the Cauchy-Riemann equations tell you if $f(z)$ is differentiable? What do they tell you if $f(\bar z) =u(x,-y)+iv(x,-y)$ is differentiable?

Edit: In view of the comments below, let me give you an example of an infinite series of functions that is uniformly convergent and where the limit is differentiable, yet the individual terms are not differentiable.
Consider
$ \sum_{n=0}^\infty f_n(z),$
where $
f_{2k}(z) = {\rm Re}(z^k),$ and $f_{2k + 1}(z) = i {\rm Im}(z^k)
$. On the closed ball $\overline{B(0, r)}$ (for $0 < r < 1$), this series converges uniformly to $ 1 / (1 - z)$.
Yet none of the individual terms (i.e. none of the $f_n(z)$'s) are differentiable.
A: Use the identity theorem.
Let $\phi(z) = f(z)-f(\bar{z})$, note that $\phi(x) = 0$ for $x$ real and so $\phi = 0$.
Hence $f(z) = f(\bar{z})$.
Since ${f(x+it) -f(x) \over it} = {f(x-it) -f(x) \over it} $ we see
that $f'(x) = 0$ for $x$ real and so $f' = 0$.
Hence $f$ is a constant.
A: HINT: Think about it in terms of $\Bbb R^2$.
$$
\frac{\partial}{\partial z}=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\\
\frac{\partial}{\partial \bar z}=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)
$$
and
$$
f(z)=f(x+iy)=:g(x,y)\\
f(\bar z)=f(x-iy)=:h(x,y)
$$
Now your problem can be restated as:
$$
\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)h(x,y)=0
$$
iff $g$ is constant.
