$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

Divide by xy(y^{2} + 1)

$\dfrac{y(y^2 + 1) \, dx}{xy(y^2 + 1)} + \dfrac{x(y^2 - 1) \, dy}{xy(y^2 + 1)} = 0$

$\dfrac{dx}{x} + \dfrac{(y^2 - 1) \, dy}{y(y^2 + 1)} = 0$

Resolve into partial fraction

$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{A}{y} + \dfrac{By + C}{y^2 + 1}$

$y^2 - 1 = A(y^2 + 1) + (By + C)y$

Set y = 0, A = -1

Equate coefficients of y^{2}

1 = A + B

1 = -1 + B

B = 2

Equate coefficients of y

0 = 0 + C

C = 0

Hence,

$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{-1}{y} + \dfrac{2y}{y^2 + 1}$

$\dfrac{y^2 - 1}{y(y^2 + 1)} = \dfrac{2y}{y^2 + 1} - \dfrac{1}{y}$

Thus,

$\dfrac{dx}{x} + \left( \dfrac{2y}{y^2 + 1} - \dfrac{1}{y} \right) \, dy = 0$

$\displaystyle \int \dfrac{dx}{x} + \int \dfrac{2y \, dy}{y^2 + 1} - \int \dfrac{dy}{y} = 0$

$\ln x + \ln (y^2 + 1) - \ln y = \ln c$

$\ln \dfrac{x(y^2 + 1)}{y} = \ln c$

$\dfrac{x(y^2 + 1)}{y} = c$

$x(y^2 + 1) = cy$ *answer - okay*