Complex analysis, find the residue 
Find the residue of $f(z)=\frac{1}{z^2\sin z}$ at $z_0=0$

What I tried
Let $g(z)=1$ and $h(z)=z^2\sin z$, both are analytics but they have zeros of different orders then $f(z)$ don't have removable singularity point at $z_0$
Then I tried to use the fact that if $lim_{z\rightarrow z_0}(z-z_0)^kf(z)$ exists, then $f(z)$ have pole of order $k$ at $z_0$, but I'm stuck on this, because if for example I take $k=2$
$$lim_{z\rightarrow z_0}(z-0)^2\frac{1}{z^2\sin z}=lim_{z\rightarrow z_0}\frac{1}{\sin z}=\infty$$
Maybe I need to do some transformation in the function, but I can not see such a transformation.

You can always expand the function into a Laurent series using $\frac{1}{z^2}\cdot$ the Taylor series of csc(z) then the residue will be the coefficient of the term indexed  by -1.
