We are given a function $f(z) = \frac{sin(1/z)}{z^2+a^2}$. We are tasked with finding its singularities, classifying them and then finding the residues at each of the singularities. Two singularities are obvious, i.e., $z = \pm ia$. Finding the residues at each of these points is also quite trivial. However, from what I can see, there is also an essential singularity at $z=0$. I am not so clear as to what the residue at this point would be.
We can write:
$sin(1/z) = z^{-1} - z^{-3}/3! + z^{-5}/5! - z^{-7}/7! \:+\:...$
$(z+ia)^{-1} = 1/ia (1 + z/ia)^{-1} = \frac{1}{ia} (1 - (z/ia) + (z/ia)^2 - (z/ia)^3 \:+\: ...)$
$(z-ia)^{-1} = -1/ia (1 - z/ia)^{-1} = -\frac{1}{ia} (1 + (z/ia) + (z/ia)^2 + (z/ia)^3 \:+\: ...)$
When we multiply it out, I noticed that we will have an infinite number of terms with $z^{-1}$. So, I am not sure what the residue at z=0 would be.
If $a=0$ the residue is clearly $0$.
For $a\neq 0$, we have the two Laurent series \begin{align*} \sin(z^{-1})&=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} z^{-2k-1}\\ \frac1{z^2+a^2}&=\frac1{a^2}\sum_{k=0}^\infty\frac{(-1)^k}{a^{2k}}z^{2k} \end{align*} that converge on $0<\lvert z\rvert<\infty$ and $\lvert z\rvert<\lvert a\rvert$ respectively. So multiplying the series, we find on the intersection $0<\lvert z\rvert<\lvert a\rvert$ (a punctured neighbourhood of $0$), $$ \frac{\sin(z^{-1})}{z^2+a^2}=\left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} z^{-2k-1}\right)\left(\frac1{a^2}\sum_{k=0}^\infty\frac{(-1)^k}{a^{2k}}z^{2k}\right). $$ The residue at $0$ is the coefficient of $z^{-1}$, which is $$ \frac1{a^2}\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\cdot\frac{(-1)^k}{a^{2k}}=\frac1{a}\sum_{k=0}^\infty\frac{a^{-2k-1}}{(2k+1)!}=\frac{\sinh(a^{-1})}{a}. $$
Alternatively, you can check the differential $\frac{\sin(z^{-1})}{z^2+a^2}\,\mathrm{d}z=-\frac{\sin(z^{-1})}{1+a^2z^{-2}}\,\mathrm{d}(z^{-1})$ is holomorphic on a neighbourhood of $z=\infty$ in the Riemann sphere $\mathbb{C}\cup\{\infty\}$, so the sum of residues at $z=0,\pm ia$ is $0$.