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

Substitute u = -y, du = -dy.

(1+x^2u)dx + x^2(x + u)du = 0

This is exact, so we set out to find a potential function.

\phi(x, u) = \int M dx = \int (1 + x^2u)dx = x + x^3u/3 + f(u)

\phi_u(x, u) = x^3 + f'(u) = x^2(x + u)
f'(u) = 2x^3/3 + x^2u
f(u) = \int f'(u) du = \int (2x^3/3 + x^2u)du
=2x^3u/3 + x^2u^2/2 + C

\phi(x, u) = x + x^3u/3 + 2x^3u/3 + x^2u^2/2 + C
\phi(x, y) = x - x^3y + x^2y^2/2 + C