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.!