The length of the longest interval in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is 

$\pi / 3$

We have,

$f(x)=3 \sin x-4 \sin ^3 x=\sin 3 x$

We know that sin x is increasing on the interval $\left(2 n \pi-\frac{\pi}{2}, 2 n \pi+\frac{\pi}{2}\right), n \in Z$ and sin 3x is periodic with period $\frac{2 \pi}{3}$ i.e. one third of the period of sin x. Therefore, the length of the longest interval

$=\frac{1}{3}\left\{\left(2 n \pi+\frac{\pi}{2}\right)-\left(2 n \pi-\frac{\pi}{2}\right)\right\}=\frac{\pi}{3}$