Bound on $f'(i)$ for $f:\text{UHP} \to \text{ region right of a hyperbola }$ 
Denote $$H=\{z=x+iy: x,y\in \mathbb{R}, y>0\}, \quad
 \text{and}\\[12pt]V=\{z=x+iy: x^2-y^2>1, x>0\}.$$
Let $$\mathcal{S} = \{f:H\to V, \,\,f \text{ is holomorphic and }
 f(i)=2\}.$$
Find all the functions $g\in \mathcal{S}$ such that $|f'(i)|\leq
 |g'(i)|$ for all $f\in \mathcal{S}$. Compute $g'(i)$.

Ideas: Suppose we have $\varphi: \mathbb{D}\to H:0\mapsto i$ and $\psi: V\to \mathbb{D}:2\mapsto 0$. Then we can reason by Schwarz's lemma somehow, but I'm not seeing a bound on $\{f'(i)\mid f\in H\}$.
 A: Your idea is exactly right (assuming you mean $\psi$ and $\varphi$ to be conformal maps onto the corresponding 
domains). By the chain rule, the derivative of the composition $\psi\circ f\circ \varphi$ at $0$ is 
$$ \varphi'(0) \, f'(i)\, \psi'(2)$$
By the Schwarz lemma, this is at most $1$ in absolute value. Hence, 
$$|f'(i)|\le \frac{1}{|\varphi'(0)\psi'(2)|}\tag{1}$$
Equality is attained in (1) if and only if $\psi\circ f\circ \varphi$ is a rotation, which means there is real 
constant $\alpha$ such that 
$$f(z)= \psi^{-1}(e^{i\alpha} \varphi^{-1}(z)),\quad z\in H  \tag{2}$$
This   identifies the set of functions $f$ which maximize $|f'(i)|$. Of course,  without concrete   $\psi$ and $\varphi$ this is not sufficiently explicit. Also, we need $\psi$ and $\varphi$ to find the value of the maximum. 
To find  $\varphi$ is standard:
$$\varphi(z)=i\frac{1-z}{1+z}$$
As for $\psi$, note that $x^2-y^2$ is the real part of $z^2$. Thus, $z\mapsto z^2-1$ sends $V$ onto the right halfplane. 
From there it's standard:
$$\psi(z) = \frac{(z^2-1)-3}{(z^2-1)+3}$$
