Prove that $\Big|\frac{f(z)-f(w)}{f(z)-\overline{f(w)}}\Big|\le \Big|\frac{z-w}{z-\overline w}\Big|$

Let $z'=f(z)$ and $w'=f(w)$. Use a Mobius transformation $\phi$ to map conformally $\mathbb H$ to the unit disk so that $z$ maps to $0$, and $w$ maps to a point $\phi(w)=\tilde w$ of the disk. Similarly, use another Mobius transformation $\psi$ to map $z'$ to $0$ and $w'$ to a point $\psi(w')=\tilde w'$ of $\mathbb D$. Consider the composition $\psi\circ f\circ \phi^{-1}: \mathbb D \to \mathbb D$ that maps $0$ to $0$ by construction. Schwarz's lemma now applies to yield $$|\psi \circ f\circ \phi^{-1} (\zeta)| \leq |\zeta|$$ for all $\zeta \in \mathbb D$. Using $\zeta=\tilde w= \phi(w)$ we obtain $$ |\psi \circ f(w)|\leq |\phi(w)|$$ If you use the exact formulas of $\phi$ and $\psi $ then you obtain the desired $$\frac{|z'-w'|}{|z'-\overline {w'}|}\leq \frac{|z-w|}{|z-\overline w|} $$

If $f\in Aut(\mathbb H)$ then $\psi \circ f\circ \phi^{-1}$ is an autmorphism of $\mathbb D$ that maps $0$ to $0$, so it is a rotation and $|\psi \circ f\circ \phi^{-1}(\zeta)|=|\zeta|$ which implies that we have equality in the above inequality.

Conversely, if we have equality, then $|\psi \circ f\circ \phi^{-1}(\zeta)|=|\zeta|$ for $\zeta=\phi(w)$, so the equality case in Schwarz's lemma implies that $\psi \circ f\circ \phi^{-1}(\zeta)=e^{i\theta}\zeta$, and in fact $f$ is an automorphism of $\mathbb H$ as a composition of conformal maps.