I want to classify the singularities of $$ f(z)=\frac{\sin(2z)}{(z-1)^3}$$
The Taylor series is: $\sin(2z)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}$
So:
$ \frac{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!} 2^{2k+1} z^{2k+1}}{(z-1)^3}$.
How can I go on from there?
You go nowhere from that. Insted, use the fact that\begin{align}\sin(2z)&=\sin\bigl(2(z-1)+2\bigr)\\&=\sin\bigl(2(z-1)\bigr)\cos(2)+\cos\bigl(2(z-1)\bigr)\sin(2)\end{align}and expand in power series centered at $1$ the functions $\sin\bigl(2(z-1)\bigr)$ and $\cos\bigl(2(z-1)\bigr)$.