Question: I need to find the Laurent series expansion of $\dfrac{1}{z-1} - \dfrac{1}{z^2}$ about $z=i$. 

I'm not sure how to deal with the $\dfrac{1}{z^2}$

Answer: 
When $\lvert z-i\rvert>1$, write $$\frac1{z^2}=\frac1{(z-i+i)^2}=\frac1{(z-i)^2}\frac1{\Bigl(1+\dfrac i{(z-i)}\Bigr)^2}$$
and apply the (hopefully known) Taylor series of $$\frac1{(1+\zeta)^2}$$
valid for $\lvert\zeta\rvert<1$.
The taylor expansion of $z^{-2}$ at $i$ is
$$z^{-2}=\sum\_{n=0}^{\infty}\frac{(-1)^ni^{n+1}}{(n+1)!}(z-i)^n$$
for $\left|z-i\right|<1$. This can be easily shown with the $n$th derivative of $z^{-2}$. Now you just need to expand $(z-1)^{-1}$ at $i$ which I am sure you know how