$\Res_{z=0}\frac{1}{z^2\sin z}$

First expand \(\inverseof{(\sin(z))}\): \[ {1\over \sin(z)} &= \inverseof{\qty{z - {1\over 3!}z^3 + {1\over 5!}z^5 -\cdots }} \\ &= \inverseof{z} \inverseof{\qty{1 - {1\over 3!}z^2 + {1\over 5!}z^4 - \cdots }} \\ &= \inverseof{z} \qty{1 + \qty{{1\over 3!}z^2 - {1\over 5!} z^4 + \cdots} + \qty{{1\over 3!}z^2 - \cdots}^2 + \cdots } \\ &= \inverseof{z}\qty{1 + {1\over 3!}z^2 + O(z^4) } ,\] using that \(\inverseof{(1-x)} = 1 + x + x^2 + \cdots\) with \(x={1\over 3!}z^2 - {1\over 5!} z^4 + \cdots\).

Thus \[ z^{-2}\inverseof{\qty{\sin(z)}} &= z^{-2} \cdot \inverseof{z}\qty{1 + {1\over 3!}z^2 + O(z^4) } \\ &= z^{-3} + {1\over 3!}\inverseof{z} + O(z) ,\] and the residue is the coefficient of \(\inverseof{z}\): \[ \Res_{z=0} {1\over z^2 \sin(z)}=\boxed{1\over 6} .\]