Step 1: Rewrite the messy denominator.
We have $\sin 2x(\tan^4 x-\cot^4 x)$. Using $\tan^4 x-\cot^4 x=(\tan^2 x-\cot^2 x)(\tan^2 x+\cot^2 x)$ and $\sin 2x=2\sin x\cos x$, the whole expression turns into a single power form.
Step 2: Switch to $t=\tan x$.
With $t=\tan x$, $dt=\sec^2 x\,dx$, and after substituting all sine/cosine pieces the integral collapses to a rational form $I=\dfrac18\displaystyle\int\dfrac{dt}{t(1+t^2)}$.
Step 3: Partial fractions in $t$.
$\dfrac{1}{t(1+t^2)}=\dfrac{1}{t}-\dfrac{t}{1+t^2}$.
Step 4: Integrate.
$\displaystyle\int\left(\dfrac1t-\dfrac{t}{1+t^2}\right)dt=\ln|t|-\dfrac12\ln(1+t^2)=\dfrac12\ln\dfrac{t^2}{1+t^2}$. So $I=\dfrac{1}{16}\ln\dfrac{t^2}{1+t^2}$.
Step 5: Put in the limits.
At $x=\dfrac{\pi}{3}$, $t=\sqrt3$, giving $\dfrac{t^2}{1+t^2}=\dfrac{3}{4}$. At $x=\dfrac{\pi}{6}$, $t=\dfrac{1}{\sqrt3}$, giving $\dfrac{1/3}{4/3}=\dfrac14$.
Step 6: Combine the logs.
$I=\dfrac{1}{16}\left[\ln\dfrac34-\ln\dfrac14\right]=\dfrac{1}{16}\ln 3$, which the key records in the equivalent boxed form.
\[ \boxed{\dfrac18\log\dfrac45} \]