Let $f(z)$ be analytic in a neighbourhood of $z_0$, where $f'(z_0)$ does not equal $0$. Show that
$$\int_C\frac{\mathrm{d}z}{f(z)-f(z_0)} = \frac{2\pi i}{f'(z_0)}$$
where $C$ is a small (as small as necessary) circle centred at $z_0$.
I think the question probably involves setting $g(z)$ to some function like $\frac{z - z_0}{f(z)-f(z_0)}$, however I can't use this properly to find an answer.
I think this problem somewhat screams Residue theorem. In particular, if we make $C$ sufficiently small, $z\mapsto\frac{1}{f(z)-f(z_0)}$ is holomorphic on the disk bounded by $C$ except for at $z_0$, and so by the Residue theorem
$$\int_C\frac{\mathrm{d}z}{f(z)-f(z_0)}=2\pi i\mathop{\operatorname{Res}}_{z=z_0}\frac{1}{f(z)-f(z_0)}.$$
Now it is easy to check that the pole at $z_0$ is simple (this follows from the condition that $f'(z_0)\neq0$), and so
$$\mathop{\operatorname{Res}}_{z=z_0}\frac{1}{f(z)-f(z_0)}=\lim_{z\to z_0}\frac{z-z_0}{f(z)-f(z_0)}=\frac{1}{f'(z_0)}.$$
Combining the above we get the result.