Let $F(x,y,z)$ be a nonsingular homogeneous polynomial on projective plane and we want to show that $X=\{[x:y:z]:F(x,y,z)=0\}$ is a Riemann surface.
My question is about the proof of holomophic compatibility(See also the last paragraph of this page). Let $X_0=\{[x:y:z]\in X:x\ne 0\}, \phi_0=y/x$ and $X_1=\{[x:y:z]\in X:y\ne 0\}, \phi_1=z/y$. We need to show that $\phi_1\circ\phi_0^{-1}$ is holomophic on $X_0\cap X_1$. I don't understand why when we write $\phi_0^{-1}(w)=[1:w:h(w)]$, then $h(w)$ is(or can be chosen to be) holomorphic.
I know there are some sort of implicit function theorem. Write $f(w,h)=F(1,w,h)$, in order to solve $h$ as a function of $w$, we need $f_h=F_h\ne 0$, but where is this implicit condition?