I finding difficulties calculating : $res(f,0)$.
with $f(z)=\frac{1}{z^2sinz}$
I thought of the method : defining $g(z)=z^3f(z)$, since $0$ is a pole of order $3$.
then : $res(f,0)=\frac{1}{2!}g^{(2)}(0).$
but calculating the 2nd derivative is harder; I tried to find the Laurent series but I couldn't :
$$f(z)=\frac{1}{z^2(z-z^3/3!+O(z^5))}$$
On
Either way is a bit of work. The first few terms of the Laurent series can be found by long division (as you've outlined):
$$f(z) = \frac{1}{z^3} + \frac{1}{6z} + \frac{7z}{360} + \textrm{higher order terms}$$
The residue is the coefficient in front of $\displaystyle\frac{1}{z}$:
$$\textrm{Res}_{z=0} f(z)=\frac{1}{6}$$
Continuing your argument:
$$ g(z)=z^3 f(z)=f(z)=\frac{z^3}{z^2 \sin(z)}=\frac{z}{\sin(z)}=\frac{1}{h(z)} $$
$$ h(z)=\frac{\sin(z)}{z}=1 - \frac{z^2}{6} + \frac{z^4}{120} + \cdots $$ Write $$ g(z) = g_0 + g_1 z + g_2 z^2 + g_3 z^3 + \cdots $$ Then $$ 1 = g(z)h(z) = g_0 + g_1 z + (g_2 - \frac{g_0}{6}) z^2 + \cdots $$ and so $$ g_2=\frac{g_0}{6}= \frac{1}{6} $$