I am new to complex analysis. Recently, I came across the statement that if $z=x+i\,y$, then for a function $f=u(x, y)+i\,v(x, y)$ satisfying Cauchy–Riemann equations, $\frac{\partial f}{\partial \bar z}=0$. The proof is given below:
Using chain rule, $$\frac{\partial f}{\partial \bar z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \bar z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar z}=0$$ Now, $x=\frac{z+\bar z}{2}$ and $y=\frac{z-\bar z}{2i}$. This gives us that $\frac{\partial x}{\partial \bar z} = \frac{1}{2}$ and $\frac{\partial y}{\partial \bar z} =\frac{i}{2}$. Substituting these values in the equation above, we get: $$\frac{1}{2}(\frac{\partial u}{\partial x}+i\,\frac{\partial v}{\partial x})+\frac{i}{2}(\frac{\partial u}{\partial y}+i\,\frac{\partial v}{\partial y})=0$$ Some simplification gives $$(\frac{\partial u}{\partial x}-\,\frac{\partial v}{\partial y}) + i\,(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y})=0$$ This gives the Cauchy–Riemann equations. However, I am a little unsure about the step where $\frac{\partial x}{\partial \bar z}$ and $\frac{\partial y}{\partial \bar z}$ are computed as they take $\frac{\partial z}{\partial \bar z}=0$. Since $\frac{\partial \bar z}{\partial z}$ is not defined (CR eqns are not satisfied), how can we take it to be $0$? Is the proof correct?
Note that $${\partial z\over \partial\overline z}{={{\partial z\over \partial x}\over {\partial\overline z\over \partial x}}+{{\partial z\over \partial y}\over {\partial\overline z\over \partial y}}\\={1\over 1}+{i\over -i}\\=0}$$