#2x^3=(3xy+1)^2#.
Diff.ing both sides w.r.t. x,
#d/dx(2x^3)=d/dx(3xy+1)^2#.
Here, by the Chain Rule,
#"The R.H.S.="d/dx(3xy+1)^2=2(3xy+1)d/dx(3xy+1)#
#=2(3xy+1){d/dx(3xy)+d/dx(1)}#
#=2(3xy+1){3d/dx(xy)+0}#
#=2(3xy+1)[3{xd/dx(y)+yd/dx(x)}]........"[Product Rule]"#
#=2(3xy+1){3(xdy/dx+y)}#
#=6(3xy+1)(xdy/dx+y).....................(1)#
#"The L.H.S.="d/dx(2x^3)=2*3x^2=6x^2...................(2)#
From #(1) and (2)#, we have,
#cancel(6)x^2=cancel(6)(3xy+1)(xdy/dx+y)", i.e.,"#
#x^2=x(3xy+1)dy/dx+y(3xy+1)#
#:. {x^2-y(3xy+1)}/{x(3xy+1)}=dy/dx#.
Enjoy Maths.!