How does this differentiation come about ? The question is that: If $f(z)$ is analytic, show that $\frac{\partial f}{\partial \bar z} = 0$  
Now, assuming $f(z) = u + iv$
$\frac{\partial f}{\partial \bar z} = \frac{\partial}{\partial \bar z}(u + iv)$
What the book does is this:  
$$
(\frac{\partial u}{\partial x}.\frac{\partial x}{\partial \bar z} + \frac{\partial u}{\partial y}.\frac{\partial y} {\partial \bar z})
$$
for the $u$ part. Similarly for the $v$ part  
Can someone please explain how that is done ?
 A: You have a function $u = u(x,y)$ of two variables. In turn, each variable is a function of two more variables, namely $z$ and $\overline{z}$. While you cannot write down $u$, you can write down $x$ and $y$ as
$$
x = \frac{z+\overline{z}}{2}, \qquad\text{ and }\qquad y = \frac{z-\overline{z}}{2i}
$$
Now you are looking to find $\frac{\partial u}{\partial \overline{z}}$. By the Chain rule, you get
$$
\frac{\partial u}{\partial x}.\frac{\partial x}{\partial \bar z} + \frac{\partial u}{\partial y}.\frac{\partial y} {\partial \bar z}
$$
In other words, you are finding the rate of change of $u$ w.r.t $\overline{z}$ : This will depend on the rate of change of $u$ w.r.t. $x$ and $y$, which in turn change w.r.t. $\overline{z}$. The chain rule helps you compute how these rates compound.
(Note : Just think of $z$ and $\overline{z}$ as symbols here, and not as complex numbers, if that is what is confusing you!)
A: I think what they are doing is writing $u(x,y)$ as $u\left(\frac{z+\bar{z}}{2}, \frac{z-\bar{z}}{2i}\right)$. Once you write it like this, you can take $\partial_z u$ by doing what they do above. Are you familiar with the chain rule?
