I have this function here which is the solution to a PDE. I have never seen a Fourier transform of a function that is both $t$ and $\omega$ dependent. What can I do with this? $$u(x,t) = \phi(x)*\mathcal{F}^{-1}\Big[e^{-(9\omega^2+2)t}\Big]\Rightarrow \phi * \mathcal{F}^{-1}\Big[e^{-9\omega^2}e^{-2t}\Big]$$
Here are the steps which lead me to this solution: $$\textbf{PDE: } u_t = 9u_{xx} - 2u; \ -\infty<x<\infty$$ $$\textbf{IC: } u(x,0) = \phi(x)$$ where
$$\phi(x) = \begin{cases} 1, \ 0\leq x\leq 1\\ 0, \ x \not\in [0,1) \end{cases} $$
Taking the Fourier transform of the PDE and IC: $$\mathcal{F}[u_t] = 9 \mathcal{F}[u_{xx}] - 2 \mathcal{F}[u]]$$ $$\Rightarrow U_t = -9\omega^2U-2U = -(9\omega^2+2)U$$ and $$\mathcal{F}[u(x,0)] = \Phi(\omega)$$ Solving the simple ODE: $$U = \Phi(\omega)e^{-(9\omega^2+2)t}$$ $$u(x,t) = \phi(x)*\mathcal{F}^{-1}\Big[e^{-(9\omega^2+2)t}\Big]\Rightarrow \phi * \mathcal{F}^{-1}\Big[e^{-9\omega^2}e^{-2t}\Big]$$
I think that a lot of your confusion is because you use a variable $\omega$ that is quite associated to frequency, i.e. time-dependency, but your Fourier transform is done in the space coordinate.
With respect to this Fourier transform, the time variable $t$ is just a constant. So, $$ u(x,t) = \phi(x)*\mathcal{F}^{-1}\Big[e^{-(9\omega^2+2)t}\Big] = \phi * \mathcal{F}^{-1}\Big[e^{-9\omega^2}e^{-2t}\Big] = \left( \phi * \mathcal{F}^{-1}\Big[e^{-9\omega^2}\Big]\right) e^{-2t} . $$