I believe it's a slightly different series apart of all the others here.
As far as my solution process goes, I'm using partial fraction decomposition first:
$$f(z) = \dfrac{1}{z\,(1-z^2)} = -\dfrac{1}{z}+\dfrac{1}{2\,(z-1)}+\dfrac{1}{2\,(z+1)}$$
and after rewriting:
$$f(z) = -\dfrac{1}{1-z+1}-\dfrac{1}{2}\,\dfrac{1}{1-z}+\dfrac{1}{2}\dfrac{1}{1--z}$$
Now coming back to the geometric series: $\displaystyle{\sum_{k=0}^{\infty}z^{k}=\dfrac{1}{1-z}}$
Based on the rewriting made easier $f(z)$ should equal $f(z) = -\displaystyle{\sum_{k=0}^{\infty}(z+1)^k+\dfrac{1}{2}\sum_{k=0}^{\infty}z^k+\dfrac{1}{2}\sum_{k=0}^{\infty}(-z)^k}$
Sadly this is wrong.
You've made a mistake while decomposing into partial fractions.. $$f(z) = \dfrac{1}{z}-\dfrac{1}{2\,(z-1)}-\dfrac{1}{2\,(z+1)}$$ Hence it can be re-written as $$\dfrac{1}{z}+\dfrac{1}{2\,(1-z)}-\dfrac{1}{2\,(1-(-z))}$$ $$=\frac{1}{z}+\frac{1}{2}\sum_{k=0}^{\infty}z^k-\frac{1}{2}\sum_{k=0}^{\infty}(-z)^k$$ Notice that when $k$ is even, the summations cancel each other. Hence, $$f(z)=\sum_{k=-1}^{\infty}z^{2k+1}$$