$\left( \frac{1+x}{1-ax}\right)^{\frac{1}{3}}$
I wish to find the Laurent series about the singular point $x=1/a$. I can find an expansion for the left side ($x=0$) and the right side ($x \rightarrow \infty$) but I don't see how to expand about the singular point.
I assume $a \neq -1$ or else this is easy.
There is no neighbourhood $U \subset \Bbb C$ of $a^{-1}$ such that $f(x)$ is holomorphic on $U \setminus \{a^{-1} \}$, so what you are asking for is impossible.
However, if you instead use the variable $t=(x-a^{-1})^{1/3}$, then you can do something (I will assume all the cube roots are the real cube roots and not some complex cube root).
So replacing $x$ with $a^{-1}+t^3$, we have $f(x) = (1+x)^{1/3}(1-ax)^{-1/3} = (1+a^{-1}+t^3)^{1/3}(-t^3)^{-1/3} = -(1+a^{-1}+t^3)^{1/3}t^{-1} = \\ -(1+a^{-1})^{1/3}t^{-1} - \frac 13 (1+a^{-1})^{-2/3}t^2 + \frac 19(1+a^{-1})^{-5/3}t^5 - \frac 5 {27}(1+a^{-1})^{-8/3}t^8 + \dots$
(if you work over $\Bbb C$ you may have to multiply this by a cube root of $1$ depending on the branches of $f$ and of $t$ you are using)