The complex version of the chain rule

The question is taken in the context of Wirtinger Derivatives.

To that end, we let $g$ and $f$ be functions of both $z$ and $\bar z$. Then, the composite function $g\circ f$ can be expressed as

$$g\circ f=g(f(z,\bar z),\bar f(z,\bar z))$$

The partial derivative of $g\circ f$ with respect to $z$ is then given by

$$\begin{align} \frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\ &=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\ &=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z} \end{align}$$

You do some mistakes. Note that $$\frac{\partial}{\partial x} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial x})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial x})$$ and $$\frac{\partial}{\partial y} (g \circ f) = (\frac{\partial g}{\partial x} \circ f)(\operatorname{Re}\frac{\partial f}{\partial y})+(\frac{\partial g}{\partial y} \circ f)(\operatorname{Im}\frac{\partial f}{\partial y}).$$ Now you should be able to finish your calculations.