Step 1: Decompose $f(z)=\dfrac{1}{z(1-z)(1+z)}$ into partial fractions: $f(z)=\dfrac{A}{z}+\dfrac{B}{1-z}+\dfrac{C}{1+z}$.
Step 2: Clearing denominators, $1=A(1-z)(1+z)+Bz(1+z)+Cz(1-z)$. Setting $z=0$ gives $A=1$; setting $z=1$ gives $B=\dfrac{1}{2}$; setting $z=-1$ gives $C=-\dfrac{1}{2}$. So $f(z)=\dfrac{1}{z}-\dfrac{1/2}{z-1}-\dfrac{1/2}{z+1}$.
Step 3: Integrating term by term over $C$, each term contributes $2\pi i$ times the winding number of $C$ about the corresponding pole: $\int_C f\,dz = 2\pi i\, n_0 - \pi i\, n_1 - \pi i\, n_{-1}$, where $n_0,n_1,n_{-1}$ are the winding numbers of $C$ about $0,1,-1$.
Step 4: Since $C$ is a simple closed (Jordan) curve, every point off the curve has winding number either $0$ or a fixed $\sigma=\pm1$ depending on orientation, the same for every point inside $C$. Taking $\sigma=1$ without loss of generality, each of $n_0,n_1,n_{-1}$ independently equals $0$ or $1$ according to whether that point lies inside $C$.
Step 5: Substituting all $8$ combinations of $n_0,n_1,n_{-1}\in\{0,1\}$ into $2\pi i\,n_0-\pi i\,n_1-\pi i\,n_{-1}$ reproduces exactly the values $0,\ \pi i,\ -\pi i,\ 2\pi i,\ -2\pi i$.\[\boxed{\{0,\ \pm i\pi,\ \pm 2i\pi\}}\]