holomorphic and antiholomorpic function is constant A function $f\colon\mathbb{C}\rightarrow\mathbb{C}$ is called antiholomorphic, when $f$ is real differentiable and  $\frac{\partial f}{\partial z}=0$. 
Why is a function which is  holomorphic and antiholomorphic a constant function ? 

My idea: $f$ is holomorphic. That means  $\frac{\partial f}{\partial\overline z}=\frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y})=0$
$f$ is antiholomorphic. That means $\frac{\partial f}{\partial z}=\frac{1}{2}(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y})=0$
All in all $\frac{\partial f}{\partial\overline z}=\frac{\partial f}{\partial z}\Leftrightarrow \frac{\partial f}{\partial y}=0$. 
Has anyone an idea how to continue ? 

It's actually very simple once you see it!
You showed that
$$ \frac{\partial f}{\partial z} = \frac{\partial f}{\partial \bar z} = 0 \implies \frac{\partial f}{\partial z} = \frac{\partial f}{\partial \bar z} \implies \frac{\partial f}{\partial y} = 0.$$
But by almost identical logic, you could also have shown that
$$ \frac{\partial f}{\partial z} = \frac{\partial f}{\partial \bar z} = 0 \implies \frac{\partial f}{\partial z} = - \frac{\partial f}{\partial \bar z} \implies \frac{\partial f}{\partial x} = 0.$$
Hopefully you can finish off from here!
