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.!