I looked up the Laplace transform table on tutorial.math.lamar.edu, and it says that $$t^n f(t) = (-1)^n F^n(s)$$
In my head that should equal
$$-1^2 (s/s^2 + \omega^2)^2$$
I put the question into Symbolab to check myself and it says that the answer is $$\frac{-(2s(-s^2+3\omega^2)}{(s^2+\omega^2)^3}$$
According to symbolab you have to take the derivative of the $(\mathcal{L}f)(s)$.Any guidance is much appreciated.
Well, as you stated we know that:
$$\mathscr{L}_x\left[x^\text{n}\cdot\text{f}\left(x\right)\right]_{\left(\text{s}\right)}=\left(-1\right)^\text{n}\cdot\frac{\text{d}^\text{n}}{\text{ds}^\text{n}}\left(\mathscr{L}_x\left[\text{f}\left(x\right)\right]_{\left(\text{s}\right)}\right)\tag1$$
And we also know that:
$$\mathscr{L}_x\left[\cos\left(\omega x\right)\right]_{\left(\text{s}\right)}=\frac{\text{s}}{\text{s}^2+\omega^2}\tag2$$
So, for your problem we get:
$$\mathscr{L}_x\left[x^2\cos\left(\omega x\right)\right]_{\left(\text{s}\right)}=\left(-1\right)^2\cdot\frac{\partial^2}{\partial\text{s}^2}\left(\frac{\text{s}}{\text{s}^2+\omega^2}\right)=\frac{2\text{s}\left(\text{s}^2-3\omega^2\right)}{\left(\text{s}^2+\omega^2\right)^3}\tag3$$