Title says it all.
I'm noticing a trend of failure on Mathematica/WolframAlpha's parts when trying to compute either the Laurent expansions or the residues of functions like $\sin\dfrac{1}{z}$ about $z=0$. (Specifying $z=0$ for the residue query returns nothing.)
Is there something about essential singularities $z_0$ that prevents one from computing a series expansion about or residue at $z_0$? Is this "something" explicitly mentioned in the usual definitions of series/residues that I'm missing? Admittedly, my complex analysis knowledge has started to rust.
Recall the definition of an essential singularity. If a function $f$ has an essential singularity at $z=z_0$, then
$$\lim_{z \to z_0} (z-z_0)^m f(z) = \infty$$
That is, an essential singularity is "more singular" than any order pole.
That all said, the definition of a residue still applies as the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion of $f$.
In your example of $f(z) = \sin{\left ( \frac{z}{z-1} \right )}$, there is an essential singularity at $z=1$. The Laurent expansion of $f$ about $z=1$ may be easily determined by expressing $f$ as follows:
$$\begin{align}f(z) &= \sin{\left (1+\frac1{z-1} \right )} = \sin{1} \cos{\left ( \frac{1}{z-1} \right )}+ \cos{1} \sin{\left ( \frac{1}{z-1} \right )}\\ &= \sin{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!(z-1)^{2 n}} + \cos{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)!(z-1)^{2 n+1}}\end{align}$$
Hopefully it is clear that the residue of $f$ at $z=1$ is $\cos{1}$.