Given $a>0$, $\frac{1}{x^2+a^2}$ is not a Schwarz function. Given $a>0$,  $f(x) = \frac{1}{x^2+a^2}$ is not a Schwarz function.
Please verify if this is correct:
Although Poisson Sumation formula is working for this function $f$, I think it is not Schwarz, since $f’(x)=-2x(x^2+a^2)^{-2}$, and $\lim_{x \to \infty} |-2x^5(x^2+a^2)^{-2}| = \infty$, so for $\alpha = 4$ and $\beta = 1$ there cannot exist $C_{\alpha,\beta}$ such that $|x^{\alpha}f^{(\beta)}(x)|\leq C_{\alpha,\beta}$.
Also, in Schwarz space can it be taken $\alpha, \beta = 0$?

The Schwartz space on $\mathbb{R}$ is the set of all smooth functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all nonnegative integers $\alpha$ and $\beta$, the function $x\mapsto x^{\alpha}f^{(\beta)}(x)$ is bounded.
Hint: In your case, take $\alpha=3$ and $\beta=0$ to get
$$
x^{3}f^{(0)}(x)=\frac{x^{3}}{x^{2}+a^{2}}.
$$
What can you say about the above as $|x|\rightarrow \infty$?
