I want to prove the inverse function theorem for complex functions using a planar mapping approach. All that's left for me to do is to show that: $$ g(w) = \xi(u,v)+i\ \eta(u,v) $$ satisfies the C-R equations - $\frac{\partial\xi}{\partial u}=\frac{\partial\eta}{\partial v}$ and $\frac{\partial\xi}{\partial v}=-\frac{\partial\eta}{\partial v}$. I know that $f(u,v)=u(x,y)+ i\ v(x,y)$ is a holomorphic function and so it satisfies the C-R equations, but I'm having trouble seeing how to apply that. Any help would be appreciated.
EDIT: I tried taking the derivative of $g$ with respect to $w$ using the fact that $w = f(z) \implies dw = f'(z) dz$. I then used $\ dz=dz+i\ dy\ \ $ and after a bunch of substitutions, I found that: $$ \frac{dg}{dw}=\frac{1}{f'(z)}[(\frac{\partial u}{\partial x}+i\frac{\partial u}{\partial y})(\frac{\partial \xi}{\partial u}-\frac{\partial \eta}{\partial v})+(-\frac{\partial u}{\partial y}+i\frac{\partial u}{\partial x})(\frac{\partial \xi}{\partial v}+\frac{\partial \eta}{\partial u})] $$
I have the condition that $f'(z)\neq 0$, so I think by setting this equal to zero I can recover the equalities that I want but I'm not certain as to why I would have to set it equal to zero or if that would even work.
EDIT: Setting this equal to zero, I can then show that this reduces to: $$ (\frac{\partial \xi}{\partial u}-\frac{\partial \eta}{\partial v})=-i\cdot (\frac{\partial \xi}{\partial v}+\frac{\partial \eta}{\partial u}) $$ Now since the LHS is purely real-number and the RHS is a purely-imaginary number, they can only be equal if they are both $0$. Thus, setting both sides equal to zero, we have the Cauchy Riemann equations hold for $g$ and so it is analytic (have already shown continuity). The only thing I'm still not convinced about is why I had to set $\frac{dg}{dw}=0$ in order to find this. Thanks for any insight!