Use

\[ \sin2x=2\sin x\cos x. \]

Then

\[ 3\sin x\sin2x=3\sin x(2\sin x\cos x)=6\sin^2x\cos x. \]

So

\[ \int 3\sin x\sin2x\,dx=\int 6\sin^2x\cos x\,dx. \]

Let \(u=\sin x\), so \(du=\cos x\,dx\).

Then

\[ \int 6\sin^2x\cos x\,dx=\int 6u^2\,du=2u^3=2\sin^3x. \]

Therefore

\[ \int_{\frac14\pi}^{\frac13\pi} 3\sin x\sin2x\,dx =\left[2\sin^3x\right]_{\frac14\pi}^{\frac13\pi}. \]

\[ =2\left(\frac{\sqrt3}{2}\right)^3-2\left(\frac{\sqrt2}{2}\right)^3. \]

\[ =\frac{3\sqrt3}{4}-\frac{\sqrt2}{2}. \]

Answer: \(\frac34\sqrt3-\frac12\sqrt2\).