So I have the function $$f:f(t)=\frac{4}{t^2+9}$$ and I know that
\begin{align*} \mathcal{F}\{e^{-|t|}\}(\omega) &= \frac{2}{\omega^2+1}. \end{align*}
I'd gotten proof of the latter Fourier Transformation in another question thread, but whenever I try to compute the earlier function with fourier transformation, it seems very hard to do at its current form. Wolfram Alpha can do it just fine, but computing it with a calculator just won't do.
Let $$ g(t) := \frac{2}{t^2+1} $$ In an answer to a previous question of yours, it was shown, using the inversion theorem, that $$ \mathcal{F}\left\{\frac{2}{t^2+1}\right\}(\omega) = 2\pi e^{-|\omega|} $$ Now note that $f(t) = \frac{2}{9} g(t/3)$ and recall the following result.
Thus \begin{align} \mathcal{F}\{f\}(\omega) &= \frac{2}{9}\cdot\left(3\cdot2\pi e^{-|3\omega|}\right)\\ &=\frac{4\pi}{3}e^{-3|\omega|} \end{align}