Find the set of orthogonal functions on the function $$\frac{x}{y}+\frac{y}{x}=C(xy)^2$$ where C is non zero.
What I tried doing was first multiplying both sides with $xy$ to get $x^2+y^2=Cx^3y^3$ and now I derived both sides to get
$$2x+2yy'=3C(x^2y^3+x^3y^2y')$$
and by grouping the parts with $y'$ we get $$y'(3Cx^3y^2-2y)=2x-3Cx^2y^3$$
here I try to eliminate C by putting $C=\frac{x^2+y^2}{x^3y^3}$ and i get $$y'(\frac{3(x^2+y^2)x^3y^2}{x^3y^3}-2y)=2x-\frac{3(x^2+y^2)x^2y^3}{x^3y^3}$$ canceling out i get $$y'(\frac{3(x^2+y^2)}{y}-2y)=2x-\frac{3(x^2+y^2)}{x}$$ Multiplying by $xy$ i get $$y'(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$ Now I know I have to replace $y'=\frac{dy}{dx}$with $-\frac{dx}{dy}$
So now I have $$\frac{-dx}{dy}(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
and i'm not sure how to go on from here.
I would really appreciate if somebody could help me from here on out. Or show me an alternate way of doing this