This is question 8 from CH2.5 in the textbook.
Let $w = z - 1$. Then $z = w+ 1$.
$$\frac{z^2}{z^2 -1} = \frac{(w+1)^2}{(w+1)^2 -1} = \frac{w^2 + 2w + 1}{w^2 + 2w}= 1+ \frac{1}{w^2 + 2w}=1 + \frac{1}{2w} \cdot \frac{1}{1 + w/2} = 1 + \frac{1}{2w} \sum_{n = 0}^{\infty} \left(- \frac{w}{2}\right)^n$$ which is valid for $|w/2| < 1$, i.e. $|z-1|<2$. Then
$$
\frac{z^2}{z^2 -1} = 1 + \frac{1}{2w}\left(1 - \frac{w}{2} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n}w^n \right)
= 1 + \frac{1}{2w} - \frac{1}{4} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n-1}w^{n-1}
= \frac{1}{2w} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} 2^{-n-2}w^{n}
= \frac{1}{2w} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{w^{n}}{2^{n+2}}
$$ Substitute $w = z-1$ back to the equation we get $$\frac{z^2}{z^2 -1} = \frac{1}{2(z-1)} +\frac{3}{4} + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{(z-1)^{n}}{2^{n+2}}$$
Residue of the function at $z = 1$ is the coefficient of $\frac{1}{z-1}$ in the Laurent series, which is $\frac{1}{2}$.
