Suppose that a trajectory of $(3x^2-y)dx+(3y^2-x)dy=0$ contains the point $(1,1)$. Show that it also contains the points $(1,-1)$, $(-1,1)$, $(0,1)$, $(1,0)$.
Is is possible to solve this problem without solving this equation? As far as I can tell, the ODE is exact, but I pretend I do not know how to solve it. If there is no such method, then is the only way is to solve it with the initial condition $y(1)=1$ and then show that it goes through the other three points?
The point is that since the left side is $d(x^3 + y^3 - x y)$, $x^3 + y^3 - x y$ is constant on a trajectory. Since $(1,1)$ is on the trajectory, that constant value is $1^3 + 1^3 - 1\cdot 1 = 1$. The curve $x^3 + y^3 - x y = 1$ is connected, and those other points are also on it.
But without finding the invariant function $x^3 + y^3 - x y$ (which amounts to solving the DE as an exact equation), I don't see any way of solving the problem.