Question:
Prove that if $f$ is differentiable at a point $z \in \mathbb{C}$ then $\overline{f(\bar{z})}$ is differentiable.
Solution:
We have $\bar{z}=x-i y$ and assume that $\bar{z} \in \mathbb{C}.$ We have that $f(\bar{z})=u(x,-y)+i v(x,-y)$ with $u(x,-y)$ and $v(x,-y)$ real differentiable. We need to show that $\overline{f(\bar{z})}=u(x,-y)-i v(x,-y)$ satisfy Cauchy Riemann equations, knowing that $f(z)=u(x, y)+i v(x, y)$ satisfies the equations. For simplicity define $$ U(x, y)=u(x,-y), \quad V(x, y)=-v(x,-y) $$ We have $$U_{x}=u_{x}, \quad V_{y}=v_{y}$$ so that $U_{x}=V_{y} .$ In the same way
$$ U_{y}=-u_{y}, \quad V_{x}=-v_{x} $$ so that $U_{y}=-V_{x}$.
I don't understand this solution.
In the second line, why $u(x,-y)$ and $v(x,-y)$ are real differentiable?
I also don't understand how the Cauchy-Riemann equations are used to get $U_{x}=V_{y} $ and $U_{y}=-V_{x}$.
Thanks for the help.