The Laurent series for $\sin z$ about $z_0 = 0$ is given by it's Taylor series about $z_0 = 0$ because sine is entire. Then, you have $z$ over the Taylor series of $\sin(z)$. Assuming you want to find the Laurent series to find the residue, I will first do everything approximately removing terms that won't contribute to the residue. i.e.
$\frac{z}{\sin^2(z)} = \frac{z}{\left(z-\frac{z^3}{3!}+\ldots\right)\left(z-\frac{z^3}{3!}+\ldots\right)} = \frac{z}{(z^2 - \frac{z^4}{3}+\ldots)} = \frac{z}{z^2\left(1-\frac{z^2}{3}+\ldots\right)} = \frac{1}{z(1-w)}$
where $1-w = 1-\frac{z^2}{3}+\ldots$. Then $w$ is a small term because we are expanding about $z_0=0$. So we can expand $1-w$,
$\frac{1}{1-w} = 1 + w + w^2 + \ldots$,
hence we find
$\frac{z}{\sin^2(z)} = \frac{1}{z}\left(1+w + w^2 + \ldots\right) = \frac{1}{z}(1+ (\frac{z^2}{3} + \ldots))$,
where everything has been valid if we care only about the residue. Hence, $Res\left(\frac{z}{\sin^2(z)} \right) = 1$.
If instead you actually do require a Laurent series, we can let $w = \frac{1}{z}$ and find the Laurent series of
$\frac{(1/w)}{\sin^2(1/w)}$ so we have
$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)}. $
Note that $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, we use this identity ($m=-n$) to find
$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{1/w}{ \left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)\left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)}$.
Next we multiply the series
$\frac{(1/w)}{ \sum_{n=-\infty}^0 \sum_{m=-\infty}^0 (-1)^{n+m}\frac{w^{2(n+m)+2}}{(2(-n)+1)!(2(-m)+1)!}}$
where I have used that $(-1)^n = (-1)^{-n}$ Now let $l=n+m$, yielding
$\frac{(1/w)}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+2}}{(2(m-l)+1)!(2(-m)+1)!}} = \frac{1}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+1}}{(2(m-l)+1)!(2(-m)+1)!}}$.
And you find the residue when $l=0$, so you find the residue to be 1, again!
For any given function, the Laurent series about a given point is unique. So you can pretty much use any method you want to find it. Using Cauchy integrals in the fashion you have described may work theoretically, but I'd imagine it would be really tedious to calculate.
The solution simply calculated the taylor series of $\sinh(z)$ and went from there.
Best Answer
The other answer has its benefits, but if you want to check your answer or simply bash out the "important" coefficients and ignore all that insignicantly small stuff, then here goes. Just note that we need to be careful about convergence.
We have that $\frac{1}{e^z-1} = \frac{1}{z(1+(z/2!+z^2/3!+...))}$ where we let $(z/2!+z^2/3!+...) = P(z)$. Then:
$$\frac{1}{e^z-1} = 1/z - P(z)/z + P(z)^2/z - P(z)^3/z + ... \\ = 1/z - 1/2 - z/3! + z/(2!)^2 - z^2/4! + 2z^2/(2!\cdot 3!) - z^2/(2!)^3 + O(z^3) \\ = 1/z - 1/2 + z/12 + z^2\cdot (1/6-1/24-1/8) + O(z^3) \\ = 1/z - 1/2 + z/12 + O(z^3).$$