The complex version of the chain rule I want to prove the following equality:
\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = (\frac{\partial g}{\partial z} \frac{\partial f}{\partial z}) + (\frac{\partial g}{\partial \bar{z}} \frac{\partial \bar{f}}{\partial z})
\end{eqnarray}
So I decide to do the following:
\begin{eqnarray}
\frac{\partial}{\partial z} (g \circ f) = \frac{1}{2}[(\frac{\partial g}{\partial x} \circ f)(\frac{\partial f}{\partial x}) + \frac{1}{i}(\frac{\partial g}{\partial y} \circ f)(\frac{\partial f}{\partial y})]
\end{eqnarray}
but the thing is that I am doing something wrong here since I don't get any conjugate function and any derivative with respect to $\bar{z}$ so Can someone help me to see where I am wrong and fix it please?
In fact I don't see what to do next, so I appreciate your help.
Thanks a lot in advance.

Edition:

What I've got so far is the following:
$$\frac{1}{2}[(\frac{\partial g}{\partial x} \circ f + \frac{\partial g}{\partial y} \circ f)\frac{\partial f}{\partial z} ]$$
but I'm still stuck.

The question is taken in the context of Wirtinger Derivatives.  
To that end, we let $g$ and $f$ be functions of both $z$ and $\bar z$.  Then, the composite function $g\circ f$ can be expressed as
$$g\circ f=g(f(z,\bar z),\bar f(z,\bar z))$$
The partial derivative of $g\circ f$ with respect to $z$ is then given by
$$\begin{align}
\frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\
&=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\
&=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z}
\end{align}$$
