So we know that if $g(z)=\frac{z-c}{1-\overline{c}z}$ $(c\in\mathbb{C})$ $|g(z)|=1$ for $|z|=1$. Does there exist a function $f(z)$ satisfies the following properties:
$f$ is analytic in some region containing $|z|\leq 1$.
The only $0$ of $f$ in $|z|\leq 1$ occurs at $1/2+i/2$, and it has order $3$.
$|f(z)|=1$ on $|z|=1$
$f'(0)=3/4$
I think the answer is negative but how will you show this?
We take $g(z)=\frac{z-c}{1-\overline{c}z}$ with $c=1/2+i/2$.
Then, $g(1/2+i/2)=0$, which is a zero of order one.
We can look now at $h(z):=(g(z))^3$.
The function $h$ satisfies (1), (2), and (3).
We can compute $h'(z)=3g^2(z)g'(z)=\frac{(3i((1+i) z-i)^2)}{(z-(1+i))^4}$ from where
$$|h'(0)|=\frac{3}{4}$$
We can define now $f(z):=\frac{\overline{h'(0)}}{|h'(0)|}h(z)$ for which we still have (1), (2), (3) and we can check that
$$f'(0)=|h'(0)|=\frac{3}{4}.$$