Square of an analytical function is analytical? Given $f \colon A \to \mathbb{C}$ defined by $f(x,y) = u(x,y) + iv(x,y)$ an analytical function, we define $\overline{A} = \{\overline{z}\ |\ z \in A\}$ and $g \colon \overline{A} \to \mathbb{C},\ h \colon \overline{A} \to \mathbb{C}$ as $g(\overline{z}) = \overline{f(z)}$ and $h(\overline{z}) = \overline{f(\overline{z})^2}$ (EDIT). My objective is proving g and h are analytical.
For the first one, we know $Re(A)=Re(\overline{A})$, so $\partial_x g = \partial_x u(x,-y) - i\partial_y v(x,-y) =\partial_x u - i\partial_x v$. Now, for $\partial_y g = -i\left(-\partial_y u+i\partial_y v\right) = i\partial_yu+\partial_yv \implies \partial_xu=\partial_y v\ \wedge\ \partial_yu = -\partial_x v $, thus complying with analiticity.
For the second one, I concluded that:
\begin{align*}
\partial_x h &=2f\partial_x f\\
&=2f\left(\partial_xu + i\partial_x v\right)\\
\end{align*}
\begin{align*}
\partial_y h &=(-2f)(-i)\partial_yf\\
&=2fi\partial_yf\\
&=2fi\left[-\partial_y u -i\partial_yv\right]\\
&=2f(-i\partial_y u +\partial_yv)\\
\end{align*}
Which implies $\partial_y v = \partial x u$ and $\partial_x v = -\partial_y u$, complying with analicity.
I'm not very sure about the $\partial_yh = -i(-2f\partial_yf)$ part. Can anyone give me feedback on it please? Thank you!
EDIT: h is defined as $h(x,-y) = \overline{f(x,-y)^2}$.

Counter-example for $h$: Let $A$ be the open unit disk so that $\overline A=A$. If $f(z)=z$ then $h(z)=\overline z ^{2}$ which is not analytic.
If $U$ and $V$ are  real and imaginary parts of $h$ then $U_x=2x$ and $V_y=-2x$, so C-R equations are not satisfied.
