Homework Problem: Complex Analysis Chain Rule My classmates and I were given that we had to verify, 
\begin{eqnarray}
\frac{\partial}{\partial z} (f \circ g) = (\frac{\partial f}{\partial z} \circ g)(\frac{\partial g}{\partial z}) + (\frac{\partial f}{\partial \bar{z}} \circ g)(\frac{\partial \bar{g}}{\partial z})
\end{eqnarray}
We are given the definition that $ \frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$. This computation is quite tedious to verify and we end up with about 32 terms that needs to cancel out. For example, if $g = s(x,y) + it(x,y)$we have that the term $\frac{\partial g}{\partial z} = \frac{\partial g}{\partial x} - i \frac{\partial g}{\partial y} = \frac{\partial s}{\partial x} + i \frac{\partial t}{\partial x} + i \frac{\partial s}{\partial y} - \frac{\partial t}{\partial y}$. So, as you can see many terms are introduced very quickly.
However, our professor said this was a simple two line proof. Is there an alternative approach that will yield the verification much more quickly?
 A: Just to expand a little bit on AlexR comment...
By the chain rule you may write the differential of $f\circ g$ as follows
\begin{eqnarray}d(f\circ g)&=&\left(\frac{\partial f}{\partial z}\circ g \right) dg+ \left(\frac{\partial f}{\partial\bar z}\circ g \right) d\bar g\\
&=&\left(\frac{\partial f}{\partial z}\circ g \right)
\left(\frac{\partial g}{\partial z}dz+\frac{\partial g}{\partial\bar z}d\bar z\right)+\left(\frac{\partial f}{\partial\bar z}\circ g \right)\left(\frac{\partial\bar g}{\partial z}dz+
\frac{\partial\bar g}{\partial\bar z}d\bar z\right)\\
&=&\left(\left(\frac{\partial f}{\partial z}\circ g \right)\frac{\partial g}{\partial z}+\left(\frac{\partial f}{\partial\bar z}\circ g\right)\frac{\partial\bar g}{\partial z}\right)dz+
\left(\left(\frac{\partial f}{\partial z}\circ g \right)\frac{\partial g}{\partial\bar z}+\left(\frac{\partial f}{\partial\bar z}\circ g\right)\frac{\partial\bar g}{\partial\bar z}\right)d\bar z
\end{eqnarray}
Identifying the "coefficients" in front of $dz$ and $d\bar z$, this gives you both $\frac{\partial(f\circ g)}{\partial z} $ and $\frac{\partial(f\circ g)}{\partial\bar z} \cdot$
