I don't have any idea where should I start for calculating Laurent expansion for the following function :
$$\frac{1}{1-z^2}$$
The thing that I've got in my notes is finding a form like the power series and then Laurent expansion can be written by putting all the $n$ inside the summation and you are OK.
I know the poles are $1,-1$ also. and the expansion should be calculated around the pole $z=1$.
$$\frac1{1-z^2}=\frac12\left(\frac1{1+z}+\frac1{1-z}\right)$$
So, for example, around the simple pole $\;z=-1\;$ :
$$\frac1{1-z^2}=\frac12\cdot\frac1{1+z}-\frac12\cdot\frac1{-2+1+z}=\frac12\cdot\frac1{1+z}+\frac14\cdot\frac1{1-\frac{1+z}2}$$
and thus for $\;\left|\frac{1+z}2\right|<1\iff|1+z|<2\;$ , we get the Laurent expansion of the function around $\;z=-1\;$ :
$$\frac1{1-z^2}=\frac1{2(1+z)}+\frac12\sum_{n=0}^\infty\frac{(1+z)^n}{2^n}$$
Now you try the expansion around $\;z=1\;$ ...