Find the laurent series on 2<|z|<3
f(z)=1/[(z^2)*(z-2)*(z+3)]
"(in powers of z)" means you find it around z = 0. So the centre is given. It helps if the whole question is posted.
I suggest writing the given function as $\displaystyle \frac{1}{z^2} \left( \frac{A}{z-2} + \frac{B}{z + 3}\right)$ where the partial fraction decomposition is left for you to finish.
Then expand each of $\displaystyle \frac{1}{z - 2}$ and $\displaystyle \frac{1}{z + 3}$ around z = 0. Then substitute etc.
eg. $\displaystyle \frac{1}{z - 2} = \frac{1}{1 - \frac{2}{z}} = 1 + r + r^2 + r^3 + ..... $ where you should recognise an infinite geometric series with $\displaystyle r = \frac{2}{z}$ (note that $\displaystyle \left| \frac{2}{z} \right| < 1$ since $\displaystyle 2 < |z| < 3$.