WolframAlpha outputs the answer as
$$-\frac14 \pi\left(-\coth(\pi) + \pi \operatorname{csch}^2(\pi)\right)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
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 $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\frac{1}{4} \pi \coth (\pi )-\frac{1}{4} \pi ^2 \text{csch}^2(\pi )}$$
