Ta có $y=\sqrt{\dfrac{5}{3}- \dfrac{1}{6}\sin^2 2x} =\sqrt{\dfrac{5}{3} -\dfrac{1}{12}(1-\cos 4x)}=\sqrt{\dfrac{19}{12}+\dfrac{1}{12}\cos 4x}$
Luôn có $-1\le \cos 4x \le 1 $
$\Leftrightarrow -\dfrac{1}{12} \le \dfrac{1}{12}\cos 4x \le \dfrac{1}{12}$
$\Leftrightarrow \dfrac{19}{12}-\dfrac{1}{12} \le \dfrac{19}{12}+\dfrac{1}{12}\cos 4x \le \dfrac{1}{12} +\dfrac{19}{12}$
Vậy $\dfrac{3}{2} \le y \le \dfrac{5}{3}$