Suppose that $L$ is a linear differential operator such that $Lu(x)=f(x)$.
Why does the following equality hold?
$$L \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}^n} \hat{y}(\omega) e^{i \omega t} d \omega=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}^n} L(-i \omega) \hat{y}(\omega) e^{i \omega t} d \omega $$
I think the result presumes that $L$ is a linear constant coefficient differential operator. When applied to $-i\omega$, $L$ appears to be treated as a polynomial in $-i\omega$, so that for example if $L=ad^2/dx^2+bd/dx+c$ we have $L(-i\omega)=a(-i\omega)^2+b(-i\omega)+c$.
In one dimension, we read in Rudin Real and Complex Analysis (2nd edition) 9.2(f) that if $f, g\in L^1$, $g(x)=-ixf(x)$, then $\hat{f}$ is differentiable and $\hat{f}{}'=\hat{g}$. If in the equation that appears in the question we have $L=d/dx$, $f=\hat{y}$, and plug in the definition of Fourier transform and the convention that for such $L$ we have $L(-i\omega)=-i\omega$, it reduces to the result just quoted. By iteration we get the equation for $L=d^n/dx^n$ (using a convention whereby $L(-i\omega)=(-i\omega)^n$ in that case), and by linearity for all linear constant coefficient differential operators.
A different version of the result, in $n$ dimensions, appears to be found in Rudin Functional Analysis 7.4(c). However, there we have instead of $L$ a polynomial $P$ and the differential operators $P(D)$ and $P(-D)$. This is much better notation in my humble opinion.