I'm reading Katok's Fuchsian Groups, and I'm confused on how the metric on the unit disk model is derived from the differential $$ds = \frac{2|dz|}{1-|z|^2}.$$
To be more specific, we first have the upper half-plane given by $\mathcal{H} = \{z \in \mathbb{C}: \text{Im}(z) > 0\}$ with the differential given by $$ds = \frac{\sqrt{dx^2+dy^2}}{y}$$ where $z = x + iy$. Using the differential, we can define a metric $\rho$ on $\mathcal{H}$ by taking the infimum of the lengths of paths between two given points. We also have the unit disk model $\mathcal{U} = \{z \in \mathbb{C}: |z| < 1\}$. The transformation $$f(z) = \frac{zi+1}{z+i}$$ is a 1-1 map of $\mathcal{H}$ onto $\mathcal{U}$. This map induces on a metric $\rho^*$ on $\mathcal{U}$ by $$\rho^*(z,w) = \rho(f^{-1}(z), f^{-1}(w)).$$ My confusion is when Katok says that because the following identity holds for $f$, $$\frac{2|f'(z)|}{1-|f(z)|^2} = \frac{1}{\text{Im}(z)}, $$ then $\rho^*$ can be identified with the metric derived from the differential $$ds = \frac{2|dz|}{1-|z|^2}.$$ I see the obvious resemblance between these expressions, but I don't know how to formally connect them. How do I use that identity on $f$ to show that the metric derived from that differential coincides with the metric $\rho^*$.
Edit: Fixed the denominator of the differential on the upper half-plane.
It might help to see the proof by using the following notation instead
$$f(z) \equiv z_{\mathcal{U}}(z_{\mathcal{H}})$$
where $z_S$ denotes which set the $z$'s lie in. From here we can see that
$$\frac{2\left|\frac{dz_{\mathcal{U}}}{dz_{\mathcal{H}}}\right|}{1-|z_{\mathcal{U}}|^2} = \frac{1}{\operatorname{Im}(z_{\mathcal{H}})} \implies \frac{2\left|dz_{\mathcal{U}}\right|}{1-|z_{\mathcal{U}}|^2} = \frac{|dz_{\mathcal{H}}|}{\operatorname{Im}(z_{\mathcal{H}})} \equiv ds$$
by the chain rule.