Let $\varphi:U \to U $ be analytic with $\varphi'(a) \neq 0$, show that for any analytic function $f$ on $U \smallsetminus \{\varphi(a)\}$ that $$\operatorname{res}(f, \varphi(a)) = \operatorname{res}((f \circ \varphi) \varphi',a) $$
I have noticed that the form seems similar to when we integrate along a curve $C$ parameterised by some $g:[a,b] \to \mathbb{C}$: $$\int_C f(z) \,dz = \int_a^bf(g(t))g'(t) \,dt $$
But can't actually get this to work in this case: $$\operatorname{res}(f, \varphi(a)) := \frac{1}{2\pi i}\int_{\delta B_\varepsilon(\varphi(a))} f(z) \,dz $$
The boundary is dependent on $\varphi(a)$, but the curve is not parameterised by this?
I'm not really sure how to go further?
Does $\varphi$ being analytic mean that since $\varepsilon$ is small that the ball around $\varphi(a)$ is still "ball shaped" and now it's boundary is parametersied by $\varphi(B_\varepsilon (a))$, or is there some better approach?