Showing that $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(n-\alpha)^2}=\pi^2\csc \pi \alpha \cot \pi \alpha$

In $(3)$ from this answer, it is shown that $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+z}=\pi\csc(\pi z)\tag{1} $$ Substituting $z\mapsto-z$ in $(1)$ and taking the derivative yields $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(k-z)^2}=\pi^2\csc(\pi z)\cot(\pi z)\tag{2} $$

The OP had asked, "Are there any other ways of proving it?"

We begin by writing the Fourier series,

$$\cos(\alpha x)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag1$$

for $x\in [-\pi/\pi]$. The Fourier coefficients in $(1)$ are given by

$$\begin{align} a_n&=\frac{2}{\pi}\int_0^\pi \cos(\alpha x)\cos(nx)\,dx\\\\ &=\frac1\pi (-1)^n \sin(\pi \alpha)\left(\frac{1}{\alpha +n}+\frac{1}{\alpha -n}\right)\tag2 \end{align}$$

Substituting $2$ into $1$, dividing by $\sin(\pi y)$, and setting $x=0$ n $(2)$ reveals

$$\begin{align} \pi \csc(\pi \alpha)&=\frac1\alpha +\sum_{n=1}^\infty (-1)^n\left(\frac{1}{\alpha -n}+\frac{1}{\alpha +n}\right)\\\\ &=\sum_{n=-\infty}^\infty \frac{(-1)^n}{\alpha-n}\tag3 \end{align}$$

Now differentiating with respect to $\alpha$, enforcing the substitution $\alpha\to \alpha/\pi$, and dividing by $\pi^2$ yields the coveted result

$$\csc(\alpha)\cot(\alpha)=\sum_{n=-\infty}^\infty \frac{(-1)^n}{(n-\alpha)^2}$$