$y=-\sin x.\tan3x$
$y'=-[(\sin x)'\tan3x+\sin x.(\tan3x)']$
$=-\cos x.\tan3x-\sin x.\dfrac{3}{\cos^23x}$
$=-\tan3x.\cos x-\dfrac{3\sin x}{\cos^23x}$
$=-\dfrac{\sin3x}{\cos3x}.\cos x-\dfrac{3\sin x}{\cos^23x}$
$=\dfrac{-\sin3x.\cos3x.\cos x-3\sin x}{\cos^23x}$
$=\dfrac{\dfrac{-1}{2}\sin6x\cos x-3\sin x}{\cos^23x}$
$=\dfrac{-\sin6x\cos x-6\sin x}{2\cos^23x}$
