Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = \frac{\pi}{a} \coth(\pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\pi\left[\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) + \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right)\right].$$
Computing the residues:
$$\operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},ia\right) = \lim_{z\rightarrow ia}\frac{(z-ia)\cot(\pi z)}{(z-ia)(z+ia)} = \frac{\cot(\pi ia)}{2i a} $$
and
$$ \operatorname{Res}\left(\frac{\cot(\pi z)}{z^{2}+a^{2}},-ia\right) = \lim_{z\rightarrow -ia}\frac{(z+ia)\cot(\pi z)}{(z+ia)(z-ia)} = \frac{\cot(i\pi a)}{2ia}.$$
Therefore, summing these we get
$$\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}} = -\frac{\pi\cot(i\pi a)}{ia} = \frac{\pi \coth(\pi a)}{a}.$$
You should be able to extend this idea to your sum with some effort.
This is similar to the difference between integral and Cauchy principal value
For example you know that $$\int _ {\mathbb R} x dx$$ does not exists, but
$$\lim_{R \to \infty} \int_{-R}^R x dx = 0$$
which is the Cauchy principal value.
The main point is that $\int_\mathbb R$ is not defined as a limit $\lim_{R \to \infty} \int_{-R}^R$.. For the riemann integral it is defined as
$$\lim_{a \to -\infty} \int_a^0 x dx + \lim_{b \to +\infty} \int_{0}^b x dx$$
and you can already see that neither of those converge.
A similar difference is there with the Lebesgue integral; The integral of a function $f(x)$ is defined as
$$\int f(x) dx = \int f(x) ^+ dx - \int f(x)^-dx$$ where $f(x)^+ = \max(f(x), 0)$ and $f(x)^- =- \min(f(x), 0)$. The main point is that we want the two separate integral to exists, and only then (if both exists) we sum them up. In this way there can't be any cancellation like the one happening with Cauchy principal value.
Your case is clearly in the same spirit, although it depends on how you define the double series $\sum_{-\infty}^{\infty}$. This is either defined as a double limit (in the "spirit" of the riemann integral) or as an integral with respect to the counting measure on $\mathbb Z$ (so basically it's a Lebesgue integral)
Best Answer
First, we note that $$\frac{n^2}{(1+n^2)^2}=\color{blue}{\frac1{1+n^2}}-\color{orange} {\frac{1}{(1+n^2)^2}}.$$
So $$\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}-\color{orange}{\sum_{n=1}^\infty \frac{1}{(1+n^2)^2}}.$$
Using the identities $$\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}=\frac12\pi\coth(\pi)-\frac12$$ from Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$ and $$\color{orange}{\sum_{n=1}^\infty \frac{1}{(1+n^2)^2}}=-\frac{1}{2}+\frac{1}{4} \pi \coth (\pi )+\frac{1}{4} \pi ^2 \text{csch}^2(\pi )$$ from How to calculate $\sum_{k=1}^{\infty}\frac{1}{(a^2+k^2)^2}$ after calculating $\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}$ using Parseval identity?, we reach the final result $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\frac{1}{4} \pi \coth (\pi )-\frac{1}{4} \pi ^2 \text{csch}^2(\pi )}$$