I have to classify the singularities in $\mathbb C\cup \{\infty\}$ of $$f(z)=\frac{e^{1/z}\sin(z^2)}{\sin(e^{iz})}$$ and compute the residue at $z=0$.
First I studied the zeros of the denominator: $\sin(e^{iz})=0\iff e^{ie^{iz}}-e^{-ie^{iz}}=0\iff e^{-ie^{iz}}(e^{2ie^{iz}}-1)=0\iff$ $e^{2ie^{iz}}=1\iff2ie^{iz_k}=2k\pi i$, $k\in\mathbb Z\iff e^{iz_k}=k\pi\iff iz_{k,h}=\ln(|k|\pi)+h\pi i$, for $(k,h)\in\mathbb Z\setminus\{0\}\times\mathbb Z$, since we have $Arg(k\pi)\in\{0,\pi\}$ and in conclusion $z_{k,h}=h\pi-i\ln(|k| \pi)$. Numerator is zero iff $\sin(z^2)=0\iff e^{iz^2}-e^{-iz^2}=0\iff e^{2iz^2}=1\iff2iz_l^2=2l\pi i$ iff $z_l^2=l\pi$, $l\in\mathbb Z$. Hence $z_l=\begin{cases}\pm\sqrt{|l\pi|}&l\ge 0\\\pm\sqrt{|l\pi|}i&l<0\end{cases}$. Have the zeros of denominator order one?
$f(1/z)=\dfrac{e^z\sin(z^{-2})}{\sin(e^{i/z})}\implies $ denominator vanishes at $i/z_{k,h}=\ln(|k|\pi)+h\pi i\implies$$ z_{k,h}=\dfrac{i}{\ln(|k|\pi)+h\pi i}$ hence $z=\infty$ is not an isolated sinularity. In order to classify the singularities of the function I should ckeck if there are some indices such that zeros of numerator and denominator coincide correct?
For the residue at zero I started expanding the function but it looks so complicated. I got $$\sin(e^{iz})=\sum_0^{\infty}\frac{(-)^n}{(2n+1)!}\left(e^{iz}\right)^{2n+1}$$ and $$e^{1/z}\sin(z^2)=\sum_{k\in\mathbb Z}\left(\sum_{4m+2-n=k}\frac{(-)^m}{(2m+1)!n!} \right)z^k.$$ I don't know how to find a result beacuse of the denominator term.
Define $g(z)=e^{1/z}$ and $h(z)=\frac{\sin(z^2)}{\sin(e^{iz})}$. We see that $f(z)=g(z)h(z)$. We note that $g(z)$ is analytic for all $z$ except $z=0$, where it has an essential singularity, we also note your analysis concerning the zeros of $h(z)$ and the poles of $h(z)$ appears correct. We note that you can verify their order by checking the zeros of the derivative and seeing they are not the same (i.e., we have simple zeros, I think).
We now also note for example that $h(z)$ is analytic at $z=0$. Since $h(z)$ is analytic at zero, in a small enough neighborhood of zero, we have,
$$f(z)=\sum_{n=0}^{\infty}\frac{h(z)}{n!z^n}$$
Hence the integral around the boundary of a sufficiently small neighborhood around zero is
$$2\pi iRes(f,0)=\int_C f(z)=\int_C \sum_{n=0}^{\infty}\frac{h(z)}{n!z^n} = \int_C \frac{h(z)}{2z^2}$$
This appears to now be calculatable.