$f(z)$ 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.

Let be $z_0$ s.t. $f'(z_0)\ne 0$. Then $f$ is locally invertible with analytic inverse (why?). Then, near $z_0:$
$$z\longmapsto f(\bar z)\longmapsto f^{-1}(f(\bar z)) = \bar z$$
would be analytic. Contradiction.
