$$f(z)=(z^2 - 1) \cot(\pi z^2)=(z+1)(z-1)\frac{\cos \pi z^2}{\sin \pi z^2}$$
$f$ has a singularity at $z$ iff $\sin \pi z^2 = 0$, this happens when $z^2 = k$ for $k \in \mathbb{Z}$.
Note that $\cos \pi z^2 = 0$ iff $z^2 = \frac{1}{2} + \mathbb{Z}$, which is mutually exclusive from zeros of $\sin \pi z^2 = 0$.
Case 1: $z = -1$
Since $(z+1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=-1$, then $f$ has a removable singularity at $z=-1$.
Case 2: $z = 0$
Since $\sin \pi z^2 = 0$ has a zero of order 2, and no other factors have zeros at $z=0$, then $f$ has a pole of order 2 at $z=0$ .
Case 3: $z = 1$
Since $(z-1)$ has a zero of order 1, $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at $z=1$, then $f$ has a removable singularity at $z=1$.
Case 4: $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$
Since $\sin \pi z^2 = 0$ has a zero of order 1, and no other factors have zeros at these values, then $f$ has a simple pole at $z = \pm\sqrt{k}$ for $k = -1, \pm2, \pm3,\cdots$.
Case 5: singularity at $\infty$
Let $$g(w) = f(1/w) = \left(\frac{1}{w}+1\right)\left(\frac{1}{w}-1\right)\frac{\cos \left( \frac{\pi}{w^2}\right)}{\sin \left(\frac{\pi}{w^2}\right)}$$
Note that $g$ has singularities at $w=0$ and $w = \pm \sqrt\frac{1}{k}$ for $k \in \mathbb{Z}$. This implies that $\forall R>0, \exists w\in\mathbb{C}, 0<|w|<R$ such that $g$ is has a singularity at $w$, and thus $g$ is not analytic in $0<|w|<R$ for any $R>0$.
Therefore, the singularity of $g$ at $z=0$ is a non-isolated singularity. And thus, the singularity of $f$ at $\infty$ is a non-isolated singularity.
