Use a direct Laurent expansion to show \[ \Res_{z=0} {1\over z-\sin(z)} = {3! \over 5\cdot 4} .\]
Note the necessity: one doesn’t know the order of the pole at zero, so it’s unclear how many derivatives to take.
Solution
Expand: \[ {1\over z - \sin(z)} &= \inverseof{z} \inverseof{\qty{1 - \inverseof{z} \sin(z) }} \\ &= \inverseof{z} \inverseof{\qty{1 - \inverseof{z}\qty{ z - {1\over 3!}z^3 + {1\over 5!} z^5 - \cdots}}}\\ &= \inverseof{z} \inverseof{\qty{1 - \qty{ 1 - {1\over 3!}z^2 + {1\over 5!} z^4 - \cdots}}} \\ &= \inverseof{z} \inverseof{\qty{{1\over 3!}z^2 - {1\over 5!}z^4 + \cdots}} \\ &= \inverseof{z} \cdot 3! z^{-2} \inverseof{\qty{1 - {1\over 5!/3!}z^2 + \cdots}} \\ &= {3!\over z^3} \qty{1 \over 1 - \qty{ {1\over 5\cdot 4}z^2 + \cdots}} \\ &= {3!\over z^3}\qty{1 + \qty{{1\over 5\cdot 4}z^2} + \qty{{1\over 5\cdot 4}z^2}^2 + \cdots} \\ &= 3! z^{-3} + {3!\over 5\cdot 4}\inverseof{z} + O(z) \\ .\]