= Solution
Let $h=f^{-1}$, $w_0=f(z_0)$ and $z=h(w)$. <Continuity> of $h$ gives $z\to z_0$ as $w\to w_0$, and injectivity gives $z\ne z_0$ when $w\ne w_0$. Therefore
$$
\frac{h(w)-h(w_0)}{w-w_0}=\left(\frac{f(z)-f(z_0)}{z-z_0}\right)^{-1}\longrightarrow\frac1{f'(z_0)}.
$$
The assumed nonvanishing <derivative> makes this limit well defined. It exists at every $w_0\in V$, so $h$ is <holomorphic>, with $\boxed{h'(w)=1/f'(h(w))}$.
Squaring maps the first <quadrant> bijectively onto the <upper half-plane>. For $\zeta$ in that half-plane, $|\zeta-i|<|\zeta+i|$, and the <Cayley transform> $(\zeta-i)/(\zeta+i)$ maps it bijectively onto the <unit disc>. Their composition is
$$
\boxed{F(z)=\frac{z^2-i}{z^2+i}.}
$$
The inverse is the unique first-<quadrant> <square root> of $i(1+w)/(1-w)$, and is <holomorphic> by the first argument. Hence $H(w)=g(F^{-1}(w))$ is <holomorphic> throughout the disc. Its <Taylor series> at zero converges for $|w|<1$, locally uniformly there. Substituting $w=F(z)$ gives the stated expansion, with $c_n=H^{(n)}(0)/n!$, convergent at every point of the <quadrant> and locally uniformly on it.
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