$y=\dfrac{\sin^3x-\cos^3x}{1+\sin x\cos x}$
$=\dfrac{(\sin x-\cos x)(1+\sin x\cos x)}{1+\sin x\cos x}$
$=\sin x-\cos x$
$\to dy=(\sin x-\cos x)'dx=(\sin x+\cos x)dx$
$ydy-\cos2x.dx$
$=(\sin x-\cos x)(\sin x+\cos x).dx-\cos2x dx$
$=(\sin^2x-\cos^2x)dx -\cos 2x.dx$
$=-\cos2x dx-\cos 2x dx$
$=-2\cos2x .dx\ne 0$
$\to$ không thể CM