Xy^2(1+x^2) dy+(1+y^2) x=0

Find its differential equation by variable separation method.

y =tan[C- arctan(x)]

The original equation is separable, so you should get:

Integral of [(1/(1 + y^2))dy] = Integral of [(1/(1 + x^2))dx]

which integrates to give:

arctan(y) = - arctan(x) + C

[where arctan(x) = tan^-1(x)]

Thus, on taking the tangent of both sides, we have

y =tan[C- arctan(x)]