Let $\gamma: I\rightarrow G$ be a differentiable path and $f:G\rightarrow \mathbb{C}$ a real differentiable function. It is to show, that for the path: $f\circ \gamma: I\rightarrow \mathbb{C}$ the following formula is true:
\displaystyle{ (f\ \circ \gamma)'(t) = f_{z}(\gamma(t))\gamma'(t)+f_{\overline{z}}(\gamma(t))\overline{\gamma ' (t)}}
and to conclude that (f\circ \gamma)'(t)= f'(\gamma(t))\gamma'(t).
I began like this:
let $f=u+iv$, it is : $\displaystyle{f_{x}= u_{x}+iv_{x}, f_{y}=u_{y}+iv_{y}, f_{z}=\frac{1}{2}(f_{x}-if_{y}), f_{z}=\frac{1}{2}(f_{x}-if_{y}), f_{\overline{z}}=\frac{1}{2}(f_{x}+if_{y})}$
then:
\begin{align} f \circ \gamma &= f(\gamma(t))= u(x(t),y(t))+iv(x(t),y(t)) \\ (f\ \circ \gamma)'(t) &= u_{x}x'(t)+u_{y}y'(t)+i(v_{x}x'(t)+v_{y}y'(t))\\ &= (u_{x}x'(t)+iv_{x}x'(t))+(u_{y}y'(t)+iv_{y}y'(t))\\ &=f_{x}x'(t) + f_{y}y'(t) \end{align}
stuck .
I don't think this is right so far because I don't see \overline{\gamma '(t)}, which should appear. Does anybody see the right way? Please do tell.