Let $f: D \to D'$ be biholomorphic map between two domains $\mathit{D}$ and $\mathit{D'}$. By considering the equation $f(f^{-1}(w)) = w$ (for $w \in \mathit{D'}$) show that $f$ is conformal.
My attempt:
By the definition of biholomorphism, both $f$ and $f^{-1}$ are holomorphic so that condition is satisfied but I am not sure how to use the given equation to show that the derivative of $f$ is always non-zero.
This answer expands @user8675309 's comment:
For all $z\in D$, let $w=f(z)$, then $z=f^{-1}(w)$, \begin{align} 1=\frac{d}{dw}w=\frac{d}{dw}f(z)=\frac{d}{dw}[f(f^{-1}(w))]\require{extpfeil}\xlongequal{\text{chain rule}}f'(f^{-1}(w))\cdot\frac{d}{dw}[f^{-1}(w)]\\ \implies f'(f^{-1}(w))\ne0\\ \implies f'(z)\ne0 \end{align} so $f'\ne0$ on $D$.
so $f$ is conformal.