Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
y = {x^3} - \sin 2x\\
 \Rightarrow y' = 3{x^2} - \left( {2x} \right)'.\cos 2x = 3{x^2} - 2\cos 2x\\
 \Rightarrow y'' = \left( {3{x^2} - 2\cos 2x} \right)' = 3.2x - 2.\left( {2x} \right)'.\left( { - \sin 2x} \right) = 6x + 4\sin 2x\\
{\left( {y' - 3{x^2}} \right)^2} + {\left( {\dfrac{{y'' - 6x}}{2}} \right)^2}\\
 = {\left( {3{x^2} - 2\cos 2x - 3{x^2}} \right)^2} + {\left( {\dfrac{{6x + 4\sin 2x - 6x}}{2}} \right)^2}\\
 = {\left( { - 2\cos 2x} \right)^2} + {\left( {2\sin 2x} \right)^2}\\
 = 4\left( {{{\cos }^2}2x + {{\sin }^2}2x} \right)\\
 = 4.1 = 4\\
2.\\
y = \sqrt {2x - {x^2}} \\
 \Rightarrow y' = \dfrac{{\left( {2x - {x^2}} \right)'}}{{2\sqrt {2x - {x^2}} }} = \dfrac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }} = \dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\
 \Rightarrow y'' = \dfrac{{\left( {1 - x} \right)'.\sqrt {2x - {x^2}}  - \sqrt {2x - {x^2}} '.\left( {1 - x} \right)}}{{2x - {x^2}}}\\
 = \dfrac{{ - 1.\sqrt {2x - {x^2}}  - \dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}.\left( {1 - x} \right)}}{{2x - {x^2}}}\\
 = \dfrac{{ - {{\sqrt {2x - {x^2}} }^2} - {{\left( {1 - x} \right)}^2}}}{{\left( {2x - {x^2}} \right)\sqrt {2x - {x^2}} }}\\
 = \dfrac{{\left( {{x^2} - 2x} \right) - \left( {{x^2} - 2x + 1} \right)}}{{\left( {2x - {x^2}} \right)\sqrt {2x - {x^2}} }}\\
 = \dfrac{{ - 1}}{{\left( {2x - {x^2}} \right)\sqrt {2x - {x^2}} }}\\
 \Rightarrow {y^3}.y'' + 1 = {\sqrt {2x - {x^2}} ^3}.\dfrac{{ - 1}}{{\left( {2x - {x^2}} \right)\sqrt {2x - {x^2}} }} + 1 =  - 1 + 1 = 0
\end{array}\)