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.
 A: 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
$$\begin{align}
\frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\
&=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\
&=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z}
\end{align}$$
A: You do some mistakes. Note that $$\frac{\partial}{\partial x} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial x})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial x})$$ and $$\frac{\partial}{\partial y} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial y})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial y}).$$ Now you should be able to finish your calculations.
A: Define $z=x+iy$, then$$\begin{cases}
   x=\frac{z+\overline{z}}{2}\\
   y=\frac{z-\overline{z}}{2i}
  \end{cases}.$$ We consider $x$ and $y$ are functions of $z$ and $\overline{z}$, thus we have
$$\frac{\partial }{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{1}{i}\frac{\partial}{\partial y} \right) ;\quad \frac{\partial }{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial}{\partial y} \right)$$which shows $z$ is exactly what we have already known. We consider $h$, $f$, $g$ all are the functions of $z$ and $\overline{z}$, then we have$$dh=\frac{\partial h}{\partial z}dz+\frac{\partial h}{\partial \overline{z}}d\overline{z}.$$Also we can express$g\circ f$ as
$$g\circ f=g(f(z,\overline{z}),\overline{f}(z,\overline{z})).$$Then,$$d(g\circ f)=\left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial z}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial z}\right)dz+
 \left( \frac{\partial g}{\partial z}\frac{\partial f}{\partial \overline{z}}+\frac{\partial g}{\partial\overline{z}}\frac{\partial \overline{f}}{\partial \overline{z}}\right)d\overline{z}. $$Compared with $dh$ and $d(g\circ f)$, the complex version of chain rule is proved.
