I think this is not a normal family since $\sin z$ is an entire function with an essential singularity at $\infty$. Hence by Great Picard, $\sin z$ can attain any value near $\infty$ infinitely many times with one possible exception. So I attempted to construct a sequence $f'(kz_0)$ such that it has no limit. Indeed, let $z_0=\pi/4$. Then $f'(kz_0)=\cos(k\pi/4)$ has no limit as $k\rightarrow\infty$. But the original function will tend to $0$ whenever $z$ is real. So the sequence to be constructed cannot be evaluated at real numbers. I know the following result:
For $f$ entire and $\{f(kz) : k \in \mathbb{C}\}$ a normal family, prove that $f$ is a polynomial.
But I wounder if there is a way to construct such a sequence explicitly, or if there is a way to show nonboundedness of the following expression
\begin{equation} \rho(f)=\dfrac{2|f'(z)|}{1+|f(z)|^2} \end{equation}
at some point $z_0$. Any hint will be appreciated.
EDIT: the reason why this family is not normal is $f_k(z)\rightarrow 0$ for $z\in \mathbb{R}$ and $f_k(z)\rightarrow \infty$ for some other points.
All functions $f_k(z) = \frac{\sin kz}{k}$ are holomorphic in the unit disk, so that any possible limit of a convergent subsequence is also holomorphic or identically $\infty$.
But $f_k(0) = 0$ and $$ |f_k(i/2)| = \frac{|\sinh(k/2)|}{k} \sim \frac{e^{k/2}}{2k} \to \infty $$ shows that both options are not possible.
You can also show that the spherical metric is not uniformly bounded on compact subsets. For each integer $k$ determine $y_k \in (0, 1)$ such that $\sinh(k y_k) = k$ and set $z_k = i y_k$. Then $$ \rho(f_k)(z_k) = \frac{|\cosh(k y_k|}{1 + (|\sinh(k y_k)|/k)^2} = \frac{\sqrt{1+k^2}}2 \to \infty \, . $$