Find Laurent series of $$f(z)=\frac{1}{z(z-3)}$$ at $z=0$ and $z=3$
So first $$\frac{1}{z(z-3)}=\frac{A}{z}+\frac{B}{z-3}$$
$$A(z-3)+Bz=1$$
$$(A+B)z-3A=1$$
So $A=-\frac{1}{3}$ and $B=\frac{1}{3}$
So we have $$f(z)=-\frac{1}{3z}+\frac{1}{3(z-3)}$$
In $z=0$ we have "problem" with $-\frac{1}{3z}$ so we only work on $\frac{1}{3(z-3)}$?
$$\frac{1}{3(z-3)}=-\frac{1}{9}\cdot \frac{1}{(1-\frac{3z}{9})}=\frac{1}{(1-\frac{z}{3})}=\sum_{n=0}^{\infty}(\frac{z}{3})^n=\sum_{n=0}^{\infty}(\frac{z^n}{3^n})$$
In $z=3$ we have "problem" with $\frac{1}{3(z-3)}$ so we only work on $-\frac{1}{3z}$?
Centering a Laurent series at $\;z=0\;$ will have to yield a series with convergence radius of at most $\;3\;$ since we have that pole at $\;z=3\;$ , so assuming $\;|z|<3\;$ we get:
$$\frac1{z(z-3)}=\frac13\left(\frac1{z-3}-\frac1z\right)=\frac13\left(-\frac13\cdot\frac1{1-\frac z3}-\frac1z\right)=$$
$$=-\frac19\left(1+\frac z3+\frac{z^2}{3^2}+\ldots+\frac3z\right)=-\frac19\left(\frac3z+\frac z3+\ldots\right)=-\frac1{3z}-\frac19\sum_{n=0}^\infty\frac{z^n}{3^n}$$
Justify the leap from first line to second one. And now around $\;z=3\;$ :
$$\frac13\left(\frac1{z-3}-\frac1{3+(z-3)}\right)=\frac13\left(\frac1{z-3}+\frac13\cdot\frac1{1+\frac{z-3}3}\right)=$$
$$=\frac13\left(\frac1{z-3}+\frac13\left(1-\frac{z-3}3+\frac{(z-3)^2}{3^2}-\ldots\right)\right)=\frac13\left(\frac1{z-3}+\sum_{n=0}^\infty\frac{(-1)^n(z-3)^n}{3^n}\right)$$