why does $\frac{df}{dz}=\frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right )$, why the $\frac{1}{2}$?

Well it starts with the general formula for differential $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.$$ In complex analysis we prefer to use $dz$ and $d\overline{z}$. So using, $dx = \dfrac{dz+d\overline{z}}{2}$ and $dy = \dfrac{dz-d\overline{z}}{2i},$ one finally gets $$df = \frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right ) dz + \frac{1}{2}\left ( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right )d\overline{z}.$$ Now it is natural to define $\frac{\partial}{\partial z}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right )$ and $\frac{\partial}{\partial \overline{z}}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right )$. Thanks to that we have the beautiful formula $$df = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \overline{z}} d\overline{z}.$$