How can I find the residue of
$$\frac{\text{sin}(\frac{1}{z})}{z^2+1}\quad ?$$
at $z=0$? In order to find this, I tried to calculate $a_{-1}$ in the Laurent series, and then arrived at $$\frac{\text{sin}(\frac{1}{z})}{z^2+1}=\frac{1}{z^2+1}\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}\cdot\left(\frac{1}{z}\right)^n,$$
but I don't know what to do next.
As you been told in the comments, you need series (Taylor's or Laurent's) around $\;z=0\;$ , and for this you need only a tiny circle around this point:
$$\frac{\sin\frac1z}{1+z^2}=\frac1{1+z^2}\cdot\sin\frac1z=\left(1-z^2+z^4-z^6+\ldots\right)\left(\frac1z-\frac1{3!z^3}+\frac1{5!z^5}-\ldots\right)\;(*)$$
The first series converges (absolutely and uniformly) for $\;|z^2|<1\iff |z|<1\;$, and the second one for any $\;|z|>0\;$, and we want the coefficient of $\;z^{-1}=\frac1z\;$ in the product series, so we multiply:
$$(*)=\ldots+\left(1\cdot\frac1z-\frac{z^2}{3!z^3}+\frac{z^4}{5!z^5}-\ldots+\right)+\ldots=\ldots+\left(1-\frac1{3!}+\frac1{5!}-\ldots\right)\frac1z+\ldots$$
and there you have your coefficient $\;c_{-1}\;$ ... Recognize now that alternating series and solve.