Đáp án:
$\begin{array}{l}
1)y = {\sin ^2}x\\
\Rightarrow y' = 2.cosx.sinx = \sin 2x\\
y' = 0\\
\Rightarrow \sin 2x = 0\\
\Rightarrow 2x = k\pi \left( {k \in Z} \right)\\
\Rightarrow x = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\\
2)y = \tan x\\
\Rightarrow y' = \dfrac{1}{{{{\cos }^2}x}}\\
\Rightarrow y' - {y^2} - 1\\
= \dfrac{1}{{{{\cos }^2}x}} - {\tan ^2}x - 1\\
= \dfrac{1}{{{{\cos }^2}x}} - 1 - {\tan ^2}x\\
= {\tan ^2}x - {\tan ^2}x\\
= 0\\
Vậy\,y' - {y^2} - 1 = 0
\end{array}$
