I'm having difficulties classifying the isolated singularities of the function
$$f(z)=\frac{z^4}{\cos(z^2)-1}.$$
The function $f$ is undefined when the denominator equal $0$, that is
$$\cos(z^2)-1=0 \iff \cos(z^2)=1 \iff z^2 = 2\pi n \iff z = \pm\sqrt{2\pi n}, \; n\in \mathbb{Z}.$$
So $f(z)$ has isolated singularities at every point $z=\pm\sqrt{2\pi n}, \; n\in \mathbb{Z}$.
Obviously I can't take the Laurent series of $f$ around every singularity to determine it's nature so how do I continue from here?
Consider the easier function $g(z)=\dfrac{z^2}{\cos z-1}$, so $f(z)=g(z^2)$.
For $g$, the singularity at the origin is removable, because $$ \lim_{z\to0}g(z)=-2 $$ In order to examine the singularities at $2k\pi$ for $k$ an integer, $k\ne0$, consider $$ \lim_{z\to2k\pi}\frac{z^2(z-2k\pi)^2}{\cos z-1}= \lim_{w\to0}\frac{(w+2k\pi)^2w^2}{\cos w-1}=-8k^2\pi^2 $$ with the obvious substitution $z=w+2k\pi$. Thus these points are poles of order $2$ (with residue $0$).
If $g$ has a pole of order $2$ at $z_0$, then $f(z)=g(z^2)$ has a pole of order $4$ at $\sqrt{z_0}$ (either determination): just substitute in the Laurent expansion.