Classifying a singularity $f(z)=\frac{\sin z}{\cos(z^3)-1}$ $$f(z)=\frac{\sin z}{\cos(z^3)-1}$$
I need to classify the singularity at $z=0$ and calculate the residue.
Apparently the method of evaluating this is through series, but that seems strange to me since I have only encountered apparently trivial cases through cauchy's theorem. What is the easiest approach to calculating the residue?

As you mention, we can understand this through series. (One could also proceed using the formula that Arthur mentions in his comment).
Firstly, we know that $\sin z = z - z^3/6 + \dots$ and $\cos(z^3) - 1 = (1-1) - z^6/2 + \ldots = -z^6/2 + \dots$. So the ratio looks like
$$ \frac{z - z^3/6}{-z^6/2} = \frac{1 - z^2/6}{-z^5/2}.$$
We've omitted higher order terms because they do not contribute here. This tells us that we have a pole of order $5$.
You also ask how to compute the residue, which is the coefficient of the degree $-1$ term. Let's show the let's use series method of getting the residue.
We now know that
$$ \frac{\sin z}{\cos z^3 - 1} = \frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots$$
This also means that
$$ \sin z = (\cos z^3 - 1)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We can now compare series expansions to arrive at the answer. In particular,
$$ z - z^3/6 + z^5/120 + \dots = (-z^6/2 + z^{12}/6 + \dots)\left(\frac{a_{-5}}{z^5} + \frac{a_{-4}}{z^4} + \dots + a_0 + a_1z + \dots\right).$$
We know that $a_{-5} = -2$ from above. Notice that $a_{-4} = a_{-2} = a_0 = 0$, which we can see by noticing that the products with $-z^6$ have degrees that don't appear on the left or by noticing that the left side is an odd function and $\cos z^3 - 1$ is even, so the Laurent expansion must be an odd function.
So how about $a_{-1}$? Notice that the only way to multiply two terms on the right to get a term like $z^5$ is to multiply $-z^6/2$ and $a_{-1}/z$. So by comparing coefficients, we see that
$$ a_{-1} \cdot \frac{-1}{2} = \frac{1}{120},$$
so that
$$a_{-1} = -\frac{1}{60}.$$
So this is the residue. *As an aside, notice that we didn't even have to compute the $a_{-3}$ term. This sort of thing can happen when we have really sparse series, like $\cos z^3 - 1$. In these cases, we can very quickly compute residues with series, especially when many things can be done quickly in your head, as in this case.*$\diamondsuit$
