Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 16

sqrt(x+y)=1+x^2y^2
dy/dxsqrt(x+y)=dy/dx(1+x^2y^2)
1/2sqrt(x+y)^(-1/2)(1+dy/dx)=x^2(2y)dy/dx+y^2(2x)
1/(2sqrt(x+y))+(dy/dx)/(2sqrt(x+y))=2x^2ydy/dx+2xy^2
dy/dx((1)/(2sqrt(x+y))-2x^2y)=2xy^2-(1)/(2sqrt(x+y))
(dy/dx)((1-4x^2ysqrt(x+y))/(2sqrt(x+y)))=(4x^2y^2sqrt(x+y)-1)/(2sqrt(x+y))
dy/dx=(4x^2y^2sqrt(x+y)-1)/(2sqrt(x+y))xx(2sqrt(x+y))/(1-4x^2ysqrt(x+y))
dy/dx=(4x^2y^2sqrt(x+y)-1)/(1-4x^2ysqrt(x+y))