The answer is "yes". We have
$$f(x)=\int_{-\omega}^\omega g(t)e^{2\pi itx}dt,$$
so
$$|f'(x)|^2\leq\left(\int_{-\omega}^\omega |t||g(t)|dt\right)^2\leq\omega^2\| g\|_2,$$
by Cauchy-Bounyakovski-Schwarz inequality. It remains to notice that $\| g\|_2=\| f\|_2$ according to Parseval .