Let $f$ be analytic on an open set $U$ and $a \in U, f'(a)\neq0$.
Prove that $\frac{1}{f'(a)}= \frac{1}{2\pi i} \int_C \frac{dz}{f(z)-f'(a)}$ .
My attempt: Since $f'(a)\neq 0$, $f$ is bijective on some neighborhood of $a$. Let $g$ be such local inverse of $f$ and $f(a)=b$. Then $\frac {1}{f'(a)}=g'(b)=\int_C \frac{g(z)dz}{(z-a)^2}$, and I got stuck here.
Is my attempt correct? Or should I seek another way to prove?
HINT:
Since $f(z)$ is analytic in a neighborhood of $a$, it can be represented by its Taylor series
$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(a)(z-a)^n}{n!}$$
Then, we can write
$$\oint_{C}\frac{1}{f(z)-f(a)}\,dz=\oint_{C}\frac{1}{f'(a)(z-a)\left(1+g(z)\right)}\,dz$$
where $g(z)$ is analytic and $g(a)=0$.
Can you proceed from here?