Show that $\frac{\partial }{\partial x}=\frac{\partial }{\partial z}+\frac{\partial }{\partial \bar{z}}$ and $\frac{\partial }{\partial y}=i\left (\frac{\partial }{\partial z}-\frac{\partial }{\partial \bar{z}} \right )$
Let $z=x+iy$ then $\bar{z}=x-iy$ from here
$\qquad\begin{align}\frac{\partial }{\partial x}&=\frac{\partial z }{\partial x}\cdot\frac{\partial }{\partial z}+\frac{\partial \bar{z}}{\partial x}\cdot\frac{\partial }{\partial \bar{z}}\\&=1\cdot\frac{\partial }{\partial z}+1\cdot\frac{\partial }{\partial \bar{z}}\\&=\frac{\partial }{\partial z}+\frac{\partial }{\partial \bar{z}}\end{align}$
and
$\qquad\begin{align}\frac{\partial }{\partial y}&=\frac{\partial z }{\partial y}\cdot\frac{\partial }{\partial z}+\frac{\partial \bar{z}}{\partial y}\cdot\frac{\partial }{\partial \bar{z}}\\&=i\cdot\frac{\partial }{\partial z}-i\cdot\frac{\partial }{\partial \bar{z}}\\&=i\cdot\left ( \frac{\partial }{\partial z} -\frac{\partial }{\partial \bar{z}}\right)\end{align}$
Is that okay or should I use the equalities $\frac{\partial }{\partial z}=\frac{1}{2}\left ( \frac{\partial }{\partial x}-i\frac{\partial }{\partial y} \right )$ and $\frac{\partial }{\partial \bar{z}}=\frac{1}{2}\left ( \frac{\partial }{\partial x} +i\frac{\partial }{\partial y}\right )$