Đáp án:
$\begin{array}{l}
y = {\sin ^2}x + \cos \left( {\dfrac{\pi }{3} - x} \right).\cos \left( {\dfrac{\pi }{3} + x} \right)\\
= {\sin ^2}x + \dfrac{1}{2}.2.\cos \left( {\dfrac{{\dfrac{{2\pi }}{3} + 2x}}{2}} \right).\cos \left( {\dfrac{{\dfrac{{2\pi }}{3} - 2x}}{2}} \right)\\
= {\sin ^2}x + \dfrac{1}{2}.\left( {\cos \dfrac{{2\pi }}{3} + \cos 2x} \right)\\
= {\sin ^2}x + \dfrac{1}{2}.\left( { - \dfrac{1}{2}} \right) + \dfrac{1}{2}\cos 2x\\
= {\sin ^2}x + \dfrac{1}{2}\cos 2x - \dfrac{1}{4}\\
\Leftrightarrow y' = 2.\left( {\sin x} \right)'.\sin x + \dfrac{1}{2}.2.\left( { - \sin 2x} \right)\\
= 2\cos x.\sin x - \sin 2x\\
= \sin 2x - \sin 2x\\
= 0
\end{array}$
