Show that $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*:=\{\overline z: z\in D\}$ using Cauchy Riemann equation. Please do not vote to close it as I want to find errors in my proof, which cannot be rectified on previously answered question.
I want a different proof using Cauchy Riemann equation.
Let $D\subset \mathbb C$ be a domain and suppose f is holomorphic on $D$.
Show that $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*:=\{\overline z: z\in D\}$.
Attempt: 
let $z= x+i y$ and $f(z)=u(x,y)+iv(x,y)$
$f$ is holomorphic on $D \Rightarrow u_x=v_y$ and $u_y=-v_x$
To show: $\overline{f(\overline{z})}$ is holomorphic on the domain $D^*$
Let $w\in D^* \Rightarrow w=\overline z$ for some $z \in D$
To show: $\overline{f(\overline{w})}$ satisfy Cauchy Riemann equation.
i.e. To Show: $\overline{f({z})}$ satisfy Cauchy Riemann equation.
$\overline{f({z})}= u(x,y)-iv(x,y)$
Let $v_1=-v$
$\overline{f({z})}= u(x,y)+iv_1(x,y)$
i.e. To show: $u_x={v_1}_y$ and $u_y=-{v_1}_x$
But $-v_y={v_1}_y$ and $-v_x=-{v_1}_x$
$\Rightarrow u_x=-v_y$ and $u_y=v_x$
which is not what I want.
Where I go wrong ?
 A: Your problem: $f(\bar z) = u(x,-y) + i v(x,-y)$.
Alternative solution: use that $f$ is locally a power series.
EDIT:
$$f(z) = f(x+iy) = u(x,y) + i v(x,y),$$
$$
\tilde u(x,y) + i\tilde v(x,y) = \tilde f(z) =
\overline{f(\overline z)} = u(x,-y) - i v(x,-y),
$$
$$\eqalign{
   \frac{\partial\tilde u}{\partial x}(x,y) = \frac{\partial u}{\partial x}(x,-y)\frac{\partial x}{\partial x} = \frac{\partial u}{\partial x}(x,-y)
&= \frac{\partial v}{\partial y}(x,-y) = \frac{\partial(-v)}{\partial y}(x,-y)\frac{\partial(-y)}{\partial y} = \frac{\partial\tilde v}{\partial y}(x,y),\cr
   \frac{\partial\tilde u}{\partial y}(x,y) = \frac{\partial u}{\partial y}(x,-y)\frac{\partial(-y)}{\partial y} = -\frac{\partial u}{\partial y}(x,-y)
&= \frac{\partial v}{\partial x}(x,-y) = -\frac{\partial(-v)}{\partial x}(x,-y)\frac{\partial x}{\partial x} = -\frac{\partial\tilde v}{\partial x}(x,y).
}$$
A: We have to prove that the function
$$g(w):=\overline{f(\bar w)}$$
is holomorphic on $D^*$. To this end fix a point $w\in D^*$ and consider a variable complex increment vector $W$ attached at $w$. Then
$$g(w+W)-g(w)=\overline{f(\bar w+\bar W)-f(\bar w)}=\overline{f'(\bar w)\bar W+o(|\bar W|)}\qquad(W\to0)\ .$$
It follows that
$$g(w+W)-g(w)=\overline{f'(\bar w)}\>W+o(|W|)\qquad(W\to0)\ ,$$
and this shows that $g$ is complex differentiable at $w$ with $g'(w)=\overline{f'(\bar w)}$.
