If xy log(x+y) = 1, prove that dy/dx = -y(x^2y +x+ y)/x(xy^2 + x + y)

Answer 1

(ydx+xdy)log(x+y)+ xy/(x+y) * (dx+dy)=0

(y+xy')log(x+y)=-xy(1+y')/(x+y)

but log(x+y)=1/xy
(y+xy')/xy=-xy((1+y')/(x+y)

(y+xy')(x+y)=-(x^2y^2)(1+y')

xy'(x+y)+y'(x^2y^2)=-y(x+y)-(xy)^2

y'= -y(x+y)-(xy)^2 ]/(x^2+xy+x^2y^2)

which reduces to what you want.