Solution: We have \[ z^3 -1 = (z -1 )(z- \omega )\left( z - \omega ^2 \right), \] where $\omega $ is a primitive third root of unity. We also note that $\omega $ and $\omega ^2$ are outside the given contour $\gamma $ and thus, the curve $\gamma $ contains only one singularity, that is, $z_0 = 1$. Therefore, \begin{align*} \frac{1}{2\pi \iota }\int \limits_\gamma \frac{1}{z^3 -1} \mathrm{d} z & = \frac{1}{2\pi \iota }\int \limits _\gamma \frac{1}{(z-1)\left( z^2 + z + 1 \right) }\mathrm{d} z \\ & = \frac{1}{2\pi \iota }\int \limits_\gamma \frac{f(z)}{z-1}\mathrm{d} z,~ f(z) = \frac{1}{z^2 + z + 1} \\ & = f(1) = \frac{1}{3}. \end{align*} Therefore, \[ \frac{1}{2\pi \iota } \int \limits_\gamma \frac{\mathrm{d} z}{z^3 - 1} = \frac{1}{3}. \]