Residue ${{(z+1)^2}\sin(1/z)}\over{(3z+1)^2}$ 
Calculate residue at $z=0$ of $\dfrac{{(z+1)^2\sin(1/z)}}{(3z+1)^2}$

It is essential singularity.So I am trying to get Laurent series expansion.Which in this case is hard because requires combing two expansions.

If you are familiar with the residue at infinity, you can reduce your problem to a simpler one. Set
$$ f(z) = \left( \frac{z+1}{3z+1} \right)^2 \sin \left( \frac{1}{z} \right). $$
The function $f$ has isolated singularities at $z = 0, -\frac{1}{3}, \infty$. Let us calculate the residue at infinity first. We have
$$ -\frac{1}{z^2} f \left( \frac{1}{z} \right) = -\frac{1}{z^2} \left( \frac{\frac{1}{z}  + 1}{\frac{3}{z} + 1} \right)^2 \sin(z) = -\frac{1}{z^2} \left( \frac{1+z}{3 + z} \right)^2 \sin(z). $$
Since $\left( \frac{1+z}{3+z} \right)^2 \sin(z)$ is holomorphic at $z = 0$, we see that the residue at infinity is zero. Since the total sum of residues is always zero, we have
$$ \operatorname{Res}(f, 0) = -\operatorname{Res} \left( f, -\frac{1}{3} 
\right) $$
so it is enough to calculate $\operatorname{Res} \left( f, -\frac{1}{3} 
\right)$. We have
$$ f(z) = \frac{(z+1)^2}{9} \sin \left( \frac{1}{z} \right) \frac{1}{ \left( z + \frac{1}{3} \right)^2} = \frac{g(z)}{\left( z + \frac{1}{3} \right)^2}$$
for $g$ which is holomorphic at $z = -\frac{1}{3}$. Hence, the residue $\operatorname{Res} \left( f, -\frac{1}{3} 
\right)$ will be $g' \left( -\frac{1}{3} \right)$ which can be calculated directly.
