Let us rewrite the given eqn. as, y=xsin2xcos(2x-y).....(1).
Keeping in mind that, (uvw)'=u'vw+uv'w+uvw',
we diff. both sides of (1) w.r.t. x, to get,
y'=1*sin2xcos(2x-y)+x(2cos2x)cos(2x-y)+xsin2x(cos(2x-y))'
Here, by Chain Rule,
(cos(2x-y))'=(-sin(2x-y))(2x-y)'=(-sin(2x-y))(2-y')=(y'-2)(sin(2x-y)
Hence,
y'=sin2xcos(2x-y)+2xcos2xcos(2x-y)+x(y'-2)sin2xsin(2x-y)
=sin2xcos(2x-y)+2xcos2xcos(2x-y)
+xy'sin2xsin(2x-y)-2xsin2xsin(2x-y)
:. y'(1-xsin2xsin(2x-y))=sin2xcos(2x-y)+2x{cos2xcos(2x-y)-sin2xsin(2x-y)}
=sin2xcos(2x-y)+2x{cos(2x+(2x-y))}
=sin2xcos(2x-y)+2xcos(4x-y)
Therefore,
y'=(sin2xcos(2x-y)+2xcos(4x-y))/(1-xsin2xsin(2x-y))
Enjoy Maths.!