Derivative of composed complex function Let $f:\mathbb{C}^3\rightarrow \mathbb{C}$ be a complex function, call $z_1,z_2,z_3$ the complex coordinates of $\mathbb{C}^3$. Let $g:\mathbb{C}\rightarrow \mathbb{R}$ be the square module function: $g(z):=z\overline{z}=(Re(z))^2+(Im(z))^2$.
I know it's a stupid question, but I'm insecure about the complex derivative $\displaystyle{\frac{\partial}{\partial z_1}(g\circ f)_{(z^0)}}$ in a point $z^0\in \mathbb{C}^3$. 
I think it's $\displaystyle{\frac{\partial g}{\partial z}(f(z^0))\cdot \frac{\partial f}{\partial z_1}(z^0)+\frac{\partial g}{\partial \overline{z}}(f(z^0))\cdot \frac{\partial f}{\partial z_1}(z^0)}=\overline{f(z^0)}\cdot \frac{\partial f}{\partial z_1}(z^0)+f(z^0)\cdot \frac{\partial f}{\partial z_1}(z^0)=2Re(f(z^0))\cdot \frac{\partial f}{\partial z_1}(z^0)$
but I'm not sure. Am I right?
 A: It's better to call $w$ the variable in $\mathbb C$, so that $f(z_0)=w_0$.
Then the correct formula is $$\frac{\partial}{\partial z_1}(g\circ f){(z_0)} = \frac{\partial g}{\partial w}(w_0) \frac{\partial f}{\partial z_1}(z_0) + \frac{\partial g}{\partial \overline w}(w_0) \frac{\partial \overline f}{\partial z_1}(z_0)=\overline w_0  \frac{\partial f}{\partial z_1}(z_0)+w_0  \frac{\partial \overline f}{\partial z_1}(z_0)                   $$ 
An example
Suppose that $f(z)=f(z_1,z_2,z_3)=z_1z_2z_3$.
 Then $(g\circ f){(z)}=z_1\bar z_1z_2\bar z_2z_3\bar z_3\;$ (I have written  $z$ instead of $z_0$, etc.)
So, the left hand side of my formula is $$\frac{\partial}{\partial z_1}(g\circ f){(z)}=\bar z_1z_2\bar z_2z_3\bar z_3 $$  The right hand side is (remember that $\frac{\partial \overline f}{\partial z_1}=\overline {\frac{\partial  f}{\partial \bar {z_1}}}=\bar 0=0$ !) $$\bar z_1\bar z_2\bar z_3.z_2z_3+   z_1\bar z_1z_2\bar z_2z_3\bar z_3.0 = \bar z_1\bar z_2\bar z_3.z_2z_3              $$  It all checks out!
