You don't need to integrate \(\displaystyle \cos(x^3)\) in order to answer this question, which is an exercise in using the Fundamental Theorem of Calculus. Roughly speaking, that theorem says that if you integrate a function and then differentiate it, you get back to where you started from.

Say that \(\displaystyle \int\cos(t^3)\,dt = F(t)\). We don't know what F(t) is, but we do know that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\). Then \(\displaystyle \frac d{dx}\int_{\sqrt[3]x}^{\pi/6}\cos(t^3)\,dt = \frac d{dx}\bigl(F(\pi/6) - F(\sqrt[3]x\,\bigr)\). You can do that differentiation by using the fact that \(\displaystyle \tfrac d{dt}F(t) = \cos(t^3)\), together with the chain rule.