\(D=R\backslash\left\{0\right\}\)
\(\sin^3x+\cos^3x=\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cos x+\cos^2x\right)=\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)\)
\(2-\sin2x=2-2\sin x\cos x=2\left(1-\sin x\cos x\right)\)
\(\Rightarrow y=\dfrac{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}{2\left(1-\sin x\cos x\right)}=\dfrac{\sin x+\cos x}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}y'=\dfrac{2\cos x-2\sin x}{4}=\dfrac{1}{2}\left(\cos x-\sin x\right)\Rightarrow y'^2=\dfrac{1}{4}\left(\cos^2x-2\sin x\cos x+\sin^2x\right)=\dfrac{1}{4}\left(1-2\sin x\cos x\right)\\y''=-\dfrac{1}{2}.\sin x-\dfrac{1}{2}\cos x\Rightarrow y''^2=\left[-\dfrac{1}{2}\left(\sin x+\cos x\right)\right]^2=\dfrac{1}{4}\left(1+2\sin x\cos x\right)\end{matrix}\right.\)
\(\Rightarrow2\left(y'^2+y''^2\right)=2\left[\dfrac{1}{4}\left(1-\sin2x\right)+\dfrac{1}{4}\left(1+\sin2x\right)\right]=1\)
