$f(z)$ is Analytic then $f(\bar z)$ is analytic iff $f$ is constant. let $f_1=u(x,y), f_2=u(x,-y)$ Then is this true that 

$(f_2)_x$ at point $(x,y) =(f_1)_x $ at point $(x,-y)$ 

And 

$(f_2)_y $ at point  $(x,y) = - (f_1)_y$ at point $(x,-y)$

If yes does this imply if  $f(z)$ is Analytic  then  $f(\bar z)$ is analytic iff $f$ is constant.
Thanks in Advance.

Without CRD:
Let $g(z):= f(\overline{z})$  and show that
$\lim_{h \to 0 , h \in \mathbb R}\frac{g(z_0+h)-g(z_0)}{h}= f'(\overline{z_0})$
and
$\lim_{h \to 0 , h \in i\mathbb R}\frac{g(z_0+h)-g(z_0)}{h}= -f'(\overline{z_0})$.
Conclusion ?
