The given equation is
#x/y^2-x=1/sin (x) + y#
differentiate both sides of the equation with respect to x
#d/dx(x/y^2-x)=d/dx(1/sin (x) + y)#
#(y^2*d/dx(x)-x*d/dx(y^2))/(y^2)^2-d/dx(x)=(sin x*d/dx(1)-1*d/dx(sin x))/(sin^2 x)+y'#
#(y^2*1-x*2y*y')/(y^4)-1=(sin x*(0)-1*(cos x))/(sin^2 x)+y'#
#(y^2-2x*y*y')/(y^4)-1=(-cos x)/(sin^2 x)+y'#
Take note: #(-cos x)/(sin^2 x)=-csc x*cot x#
#(y^2)/y^4-(2x*y*y')/(y^4)-1=-csc x*cot x+y'#
#1/y^2-(2xy')/(y^3)-1=-csc x*cot x+y'#
Using transposition, it follows
#-y'-(2xy')/(y^3)=1-csc x*cot x-1/y^2#
Multiply both sides of the equation by #y^3#
#y^3[-y'-(2xy')/(y^3)]=y^3[1-csc x*cot x-1/y^2]#
#-y^3y'-2xy'=y^3-y^3*csc x*cot x-y#
Factor the #y'# and dividing both sides by #(-y^3-2x)#
#((-y^3-2x)y')/((-y^3-2x))=(y^3-y^3*csc x*cot x-y)/(-y^3-2x)#
#cancel((-y^3-2x)y')/cancel((-y^3-2x))=(y^3-y^3*csc x*cot x-y)/(-y^3-2x)#
#y'=(y^3-y^3*csc x*cot x-y)/(-y^3-2x)#
#y'=(y^3*csc x*cot x+y-y^3)/(y^3+2x)#
God bless ....I hope the explanation is useful.
