Given the function $$f(z) = \dfrac{1}{z^2\,(z^2+1)} = \dfrac{1}{z^2}-\dfrac{1}{z^2+1}$$
Now let's span some annulus. I'm not sure with what radius.
Since $[0,\quad -i,\quad+i]$ are the zeros, I'd say it's safe to define the series on:
$\begin{array}{c}\textbf{1}.&|z|>1, &&\textbf{2}.& 0<|z|<1\end{array}$
Concerning the first array: $\displaystyle f(z) = \dfrac{1}{z^2}-\dfrac{1}{z^2}\dfrac{1}{1--\frac{1}{z^2}} = \dfrac{1}{z^2}-\dfrac{1}{z^2}\,\sum_{k = 0}^{\infty}\left(\dfrac{-1}{z^2}\right)^k$
Concerning the second array: $\displaystyle f(z) = \dfrac{1}{z^2}-\dfrac{1}{1--z^2} = \dfrac{1}{z^2}-\sum_{n = 0}^{\infty}\left(-1\,z^2\right)^k$
Are those right? Also to recap: Singularites can't be plugged into the series, so $z \neq 0 \neq-i\neq+1$.
(This should be guaranteed since the radius is restricted).
Additionally: Are those 2 functions holomorphic the reason being there are no poles within?
Lastly. Are those 2 restricted? Hence they converge?
Your Laurent series are correct, although I would rather write the first one as$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{z^{2k}}$$and the second one as$$\sum_{k=-1}^\infty(-1)^{k+1}z^{2k}.$$The first one converges if and only if $|z|>1$, whereas the second one converges if and only if $0<|z|<1$. There are no poles inside these regions because that's what always happens for Laurent series; their sum is always a holomorphic functions.