Find laurent series for $f(z)=\dfrac{1}{z^2-1}+\dfrac{1}{z(z-1)};z_0=0$ that converges in $0<|z|<1$.

I tried to find the solution for the first fraction like this.

\begin{align} \frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}&=-\frac{1}{2(z+1)}+\frac{1}{2(z-1)} \\ \end{align} The second term is $-\dfrac 12\dfrac{1}{1-z}$ and $|z|<1$ on $0<|z|<1$. So its $-\dfrac 12 \sum\limits_{k=0}^{\infty}z^k$.

Second fraction $$\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}=-\dfrac 1z-\sum\limits_{k=0}^{\infty}z^k$$

all together $$-\dfrac 1z-\underbrace{\sum\limits_{k=0}^{\infty}z^k-\dfrac 12 \sum\limits_{k=0}^{\infty}z^k}_{-\frac 32 \sum\limits_{k=0}^{\infty}z^k } -\dfrac 12\dfrac{1}{1+z}$$

Am I on the right track? What tricks are needed for term $-\dfrac 12\dfrac{1}{1+z}$?

EDIT: I had an idea... $$-\dfrac 12\dfrac{1}{1+z}= -\dfrac 12\dfrac{1}{1-(-z)}=-\frac 12 \sum \limits_{k=0}^{\infty}(-1)^nz^k$$ so the series is...(maybe) $$-\dfrac 1z-\dfrac 32 \sum\limits_{k=0}^{\infty}z^k -\frac 12 \sum \limits_{k=0}^{\infty}(-1)^nz^k$$

Powers of $z$, right? One of them being a constant times $z^{-1}$, right? Just use geometric series: $(z-1)^{-1}=-1/(1-z)=-\sum_{n\ge0}z^n$, so that the troublesome part is $-\sum_{n\ge-1}z^n$, and the easy part is $-1/(1-z^2)=-\sum_{n\ge0}z^{2n}$. Add them up and I think I’ve gotten the same thing you did.