1. Let $I=\int_{\pi/6}^{\pi/3} \tan^3 x\,\sin^2 3x\left(2\sec^2 x\,\sin^2 3x+3\tan x\,\sin 6x\right)dx.$

We look for a substitution by spotting a product-rule pattern.

2. Consider $u=\tan^2 x\,\sin^2 3x.$ Then $\frac{du}{dx}=2\tan x\sec^2 x\sin^2 3x+\tan^2 x\cdot 2\sin 3x\cdot 3\cos 3x.$ Using $2\sin 3x\cos 3x=\sin 6x$, this becomes $\frac{du}{dx}=2\tan x\sec^2 x\sin^2 3x+3\tan^2 x\sin 6x.$

3. Now factor the integrand: $\tan^3 x\sin^2 3x\left(2\sec^2 x\sin^2 3x+3\tan x\sin 6x\right)$ $=\tan^2 x\sin^2 3x\left(2\tan x\sec^2 x\sin^2 3x+3\tan^2 x\sin 6x\right)$ $=u\,\frac{du}{dx}.$ Hence $I=\int_{\pi/6}^{\pi/3} u\,\frac{du}{dx}\,dx=\int u\,du=\frac{u^2}{2}\Big|_{\pi/6}^{\pi/3}.$

4. Compute the boundary values.

At $x=\pi/3$: $\tan^2\frac{\pi}{3}=3,\qquad \sin^2(\pi)=0,$ so $u\left(\frac{\pi}{3}\right)=3\cdot 0=0.$

At $x=\pi/6$: $\tan^2\frac{\pi}{6}=\frac13,\qquad \sin^2\frac{\pi}{2}=1,$ so $u\left(\frac{\pi}{6}\right)=\frac13.$

Therefore, $I=\frac12\left[0^2-\left(\frac13\right)^2\right]=\frac12\left(-\frac19\right)=-\frac{1}{18}.$

5. Hence the correct option is $\boxed{B\;\left(-\frac{1}{18}\right)}.$