Suppose we know the Fourier transform of $f(t)$.
Is it possible to compute the Fourier transform of $\frac{f(t)}{(t^2-a^2)^2}$ in terms of the Fourier transform of f(t) ?
The rational function $\frac{1}{(t^2-a^2)^2}$ can be decomposed as $\frac{1}{4 a^3 (a+t)}+\frac{1}{4 a^2 (a+t)^2}-\frac{1}{4 a^3 (t-a)}+\frac{1}{4 a^2 (t-a)^2}$ but I'm not sure it is possible in general to find $\mathscr{F}\Big[\frac{f(t)}{a\pm t}\Big]$ or $\mathscr{F}\Big[\frac{f(t)}{(a \pm t)^2}\Big]$ in terms of the Fourier transform of $f(t)$.
The Fourier Transform of $g(t)=1/(t\pm |a|)$ is
$$\mathscr{F}\{g\}(\omega)=i\pi \text{sgn}(\omega)e^{\mp \omega |a|}$$
Then, using the convolution theorem reveals
$$\begin{align} \mathscr{F}\{fg\}(\omega)&=\mathscr{F}\{f\}*\mathscr{F}\{g\}\\\\ &=\int_{-\infty}^\infty \left(\mathscr{F}\{f\}(\omega')\right)\left(i\pi \text{sgn}(\omega')e^{\mp (\omega-\omega') |a|}\right)\,d\omega'\\\\ &=i\pi e^{\mp \omega|a|}\int_{-\infty}^\infty \left(\mathscr{F}\{f\}(\omega')\right)\left(i\pi \text{sgn}(\omega')e^{\pm \omega' |a|}\right)\,d\omega'\\\\ \end{align}$$
Is this what you sought?