Cauchy's Integral Formula states that for $a \in \mathbb{C}$, $r > 0$, if $f:B_r(a) \to \mathbb{C}$ is holomorphic, then for every $w \in B_r(a)$ such that $|w-a| < \rho < r$ we have
$$ \int_{|z-a|=\rho} \frac{f(z)}{z-w} dz = 2 \pi \textbf{i}\: f(w). $$
Here $\gamma$ is the unit circle travelled anti-clockwise, so I have taken $a = 0$ and $\rho = 1$. I see that $\frac{1}{z^2 - \frac{1}{4}} = \frac{1}{\left(z-\frac{1}{2}\right)\left(z + \frac{1}{2}\right)}$. So I have taken $f(z) = \left(z + \frac{1}{2}\right)^{-1}$, which means $w = \frac{1}{2}$.
The problem now is that $f$ is undefined at $-\frac{1}{2}$, but my open ball must include $\frac{1}{2}$ because $\rho = 1 < r$, hence $f$ is not holomorphic. A similar issue occurs when we take $f(z) = \left(z - \frac{1}{2}\right)^{-1}$. What am I missing here?
The problem is easily resolved by noting that $\frac{1}{\left(z-\frac{1}{2}\right)\left(z + \frac{1}{2}\right)}=\frac 1 {z-\frac 1 2}-\frac 1 {z+\frac 1 2}$ and applying CIF for each of the two terms.