Trying to express this as a partial fraction have proved unsuccesful, and I can't find any examples which replicate a Laurent series that isnt either easy to turn into a partial fraction, or has an obvious $\frac{1}{z}$ expression. The closest I got was by expressing the $(z^3+1)^{-1}$ term as $1 - z^3 + z^6 - z^9\dots$ (which I believe is valid, as the annulus on which $z$ is analytic requires $\mod z$ to be less than $1$) which resulted in $$\frac{1}{z} + 2 + z - z^2 - 2z^3 - z^4 + z^5 + 2z^6 +z^7\dots,$$ and I lack the brainpower to spin that into a Laurent series.
I also suck at maths, and this is part of a module that is a maths addition to my non maths course, so if you think there's a very trivial answer and I'm just being super dumb that is very much a possibility.
Thanks!
You are right, the only step left is to multiply the terms: $$\frac{(z+1)^2}{z(z^3+1)}=\frac{z^2+2z+1}{z(1-(-z)^3)}=\frac{z+2+\frac{1}{z}}{1-(-z)^3}=\left(z+2+\frac{1}{z}\right)(1+(-z)^3+(-z)^6+\ldots)$$