#y=sin(xy).#

#:. dy/dx," using the Chain Rule,"#

#=d/dx(sin(xy))={cos(xy)}{d/dx(xy)}," &, using the Product Rule,"#

#={x*d/dx(y)+y*d/dx(x)}cos(xy),#

#:. dy/dx=xcos(xy)dy/dx+ycos(xy),#

#rArr {1-xcos(xy)}dy/dx=ycos(xy).#

#:. dy/dx={ycos(xy)}/ {1-xcos(xy)}.#

**Otherwise,** #y=sin(xy) rArr arc siny=xy, or, x=(arc siny)/y.#

Hence, diff.ing both sides w.r.t. #y,# we have, by the **Quotient Rule,**

#dx/dy={y*d/dy(arc siny)-(arc siny)*d/dy(y)}/y^2,#

#={y*(1/sqrt(1-y^2))-(arc siny)*1}/y^2,#

#={y-sqrt(1-y^2)arc siny}/{y^2*sqrt(1-y^2)},#

Therefore, #dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.#

**I leave it to the Questioner to show that both Answers tally with each other.**

**Enjoy Maths.!**