So I don't know what the result is called as it is unstated in the problem but it is a proof following from the Fourier Energy theorem. Prove: $$\int_{-\infty}^{\infty}f'(x)f'^*(x)dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\omega^2\hat{f}(\omega)\hat{f}^*(\omega)d\omega$$
using: $$ \int_{-\infty}^{\infty}(f(x))^2dx = \frac{1}{2\pi}\int_{-\infty}^\infty\|\hat{f}(\omega)\|^2 $$
What I have done so far is replace $\|\hat{f}(\omega)\|^2 $ with $ \hat{f}(\omega)\hat{f}^*(\omega)$ to get
$$ \int_{-\infty}^{\infty}f(x)f(x)dx = \frac{1}{2\pi}\int_{-\infty}^\infty\hat{f}(\omega)\hat{f}^*(\omega) $$
now my idea was to use the Fourier transform property: $\mathcal{F}\left(\frac{d^nf}{dx^n}\right) = (i\omega)^n\hat{f}(\omega)$ but was confused as to how this would work, any ideas?
you nearly got it. If you use in your formula $f'$ instead of $f$ you can use $\mathcal F(f')(\omega)=i\omega f(\omega)$.
So we are using the Parval identiy on $f'$ to get: $$\int_{-\infty}^\infty f'(x)f'^{*}(x)dx=\tfrac 1 {2\pi}\int_{-\infty}^\infty\widehat{ f'}(\omega)\widehat {f'}^*(\omega)d\omega$$ Now we can use the Fourier poperty and get with $\widehat{f'}(\omega)=i\omega\hat f(\omega)$: $$\tfrac 1{2\pi}\int_{-\infty}^\infty\underbrace{(i\omega)\hat f(\omega)}_{=\widehat{f'}(\omega)}\underbrace{(i\omega)^*\hat f^*(\omega)}_{=\widehat{f'}^*(\omega)}d\omega=\tfrac 1 {2\pi}\int_{-\infty}^\infty\underbrace{-i^2}_{=1}\omega^2\hat f(\omega)\hat f^*(\omega)d\omega$$