I am reading complex number from mathematical methods point of view.
There comes Cauchy-Riemann condition in different form.
Suppose $f:\mathbb C\to\mathbb C$.
As $z=x+iy$
So, $\displaystyle f(z)=u(x,y)+iv(x,y)$
By Cauchy-Riemann condition (necessary condition for differntiability) at $z_o$:
$\boxed{\displaystyle\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\;and\\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}}\tag{1}$
As $\displaystyle x=\frac{z+z^*}{2}\;and\;y=\frac{z-z^*}{2i}$
$\displaystyle f(z)=u(\frac{z+z^*}{2},\frac{z-z^*}{2i})+iv(\frac{z+z^*}{2},\frac{z-z^*}{2i})\tag{2}$
$\displaystyle f(z,z^*)=u(\frac{z+z^*}{2},\frac{z-z^*}{2i})+iv(\frac{z+z^*}{2},\frac{z-z^*}{2i})\tag{3}$
By using chain rule for differentiation,
$\displaystyle\frac{\partial f(z,z^*)}{\partial z^*}=\Big(\frac{\partial u}{\partial x}\frac{\partial x}{\partial z^*}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial z^*}\Big)+i\Big(\frac{\partial v}{\partial x}\frac{\partial x}{\partial z^*}+\frac{\partial v}{\partial y}\frac{\partial x}{\partial z^*}\Big)$
But $\displaystyle\frac{\partial x}{\partial z^*}=\frac{1}{2}$ and $\displaystyle\frac{\partial y}{\partial z^*}=-\frac{1}{2i}$
Thus we have $\displaystyle\frac{\partial f(z,z^*)}{\partial z^*}=\Big(\frac{1}{2}\frac{\partial u}{\partial x}-\frac{1}{2i}\frac{\partial u}{\partial y}\Big)+i\Big(\frac{1}{2}\frac{\partial v}{\partial x}-\frac{1}{2i}\frac{\partial v}{\partial y}\Big)$
$\displaystyle\implies\frac{\partial f(z,z^*)}{\partial z^*}=\frac{1}{2}\Big(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\Big)+\frac{i}{2}\Big(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\Big)$
From $(1)$,
$\boxed{\displaystyle\displaystyle\frac{\partial f(z,z^*)}{\partial z^*}=0}$
This is the complex form of Cauchy-Riemann condition.
So for a complex valued function to be differetiable at a point $z_o$, it should not depend explicitly on the $z^*$.
In the proof, in $(2)$ and $(3)$, it has been assumed that $z$ and $z^*$ are independent of each other.
I have a doubt that why is it so? Basically $f(z)=u(x,y)+iv(x,y)$.
$u:\mathbb R^2\to\mathbb R$. But later we write $u(x,y)$ as $u(\frac{z+z^*}{2},\frac{z-z^*}{2i})$.
If $z$ and $z^*$ are independent of each other this means that $z$ and $z^*$ can take any values independent of each other, basically they are just two complex numbers which can take any value. So how can we be sure that $\frac{z+z^*}{2}$ is a real number, it can be complex number and thus the domain of $u$ becomes $\mathbb C^2$ instead of $\mathbb R^2$.
Though later we get as a reult that $f$ does not depend on $z^*$, but isn't that obvious because our hypothesis that $z$ and $z^*$ are independent is wrong.
I have doubt that in the proof how we can take $z$ and $z^*$ to be independent of each other? Is there any specific reason for that?
I think that I am missing something.
I think that statements like " f is holomorphic if and only if it does not depend on $z^*$" can be more misleading than enlightening at the beginning.
Indeed you are right, the expressions $x=\frac{z+z^*}2,y=\frac{z-z^*}{2i}$ do not represent a change of variable. Rather they just express real and imaginary parts in terms of the conjugation operation. Then speaking about the chain rule for differentiation is not particularly meaningful.
A pragmatic approach is to define the operators $\partial/\partial z:=\frac 12(\partial/\partial x-i\partial/\partial y)$ and $\partial/\partial z^*:=\frac 12(\partial/\partial x+i\partial/\partial y)$ acting on functions of two variables $x,y$. This is a definition; it has nothing to do with chain rule. To motivate the definition, let us also consider the following pretty natural definitions: $dz:=dx+i dy$ and $d z^* := dx-i dy$. Then one can easily prove that $$ d f = \frac{\partial f}{\partial x}d x + \frac{\partial f}{\partial y}d y =\frac{\partial f}{\partial z}d z + \frac{\partial f}{\partial z^*}d z^*. $$ Moreover, the Cauchy-Riemann can be written down as $\partial f/\partial z^* = 0$ (but this is at this stage a mere rewriting, essentially we have defined $\partial/\partial z^*$ to make this equality true).
Two more comments.
Consider a bivariate polynomial with complex coefficient $f(x,y)\in\mathbb C[x,y]$, $$ f(x,y) = \sum_{l,m} f_{l,m}x ^l y^m. $$ Then coefficients $\widehat f_{l,m}\in\mathbb C$ such that $$ f(x,y) = \sum_{l,m} \widehat f_{l,m}(x+i y) ^l (x-i y)^m. $$ are completely determined by the $f_{l,m}$'s. It is an exercise to check that indeed $F(z) := f(Re z, Im z)$ is holomorphic if and only if $\widehat f_{l,m}=0$ for all $m>0$; thus at least for polynomials we can make sense of the statement of "independence from $z^*$". One could proceed similarly with analytic functions of two variables $x,y$ (which are convergent power series in $x,y$, and now this actually covers all holomorphic functions!).
The linear differential operators $\partial/\partial z$ and $\partial/\partial z^*$ can be defined just by the properties $$ \frac{\partial}{\partial z} z = 1,\qquad \frac{\partial}{\partial z} (z^*) = 0,\qquad (\star) $$ and $$ \frac{\partial}{\partial z^*} z = 0,\qquad \frac{\partial}{\partial z^*} (z^*) = 1. $$ Indeed, if we insist that $\partial/\partial z$ should be a linear differential operator with complex coefficients, we should write $z=x+iy$ and $\frac{\partial}{\partial z} = (a+ib)\frac{\partial}{\partial x}+(c+id)\frac{\partial}{\partial y}$, and the conditions $(\star)$ imply $a-d=1,a+d=0,b=c=0$, yielding the previous definition (and similarly for the operator $\partial/\partial z^*$).