Suppose I am given a vector field $X$ over a two-dimensional Euclidean Space. Suppose that the equation $\{f=0\}$ (which defines a curve in the plane) is the equation of a orbit of the vector fields $X$.
What does this means in terms of $X(f)$ I am guessing that the set $\{f=0\}$ must be $X$-invariant, therefore $f$ satisfies $X(f)=p\cdot f$, $p$ being a constant. is this right?
EDIT: I should also add that $f$ is an irreductible polynomial! We can also consider that $X$ is a polynomial field of the form $X=px\partial_x+qy\partial_y$ where $p$ and $q$ are positive integer which are relatively primes.
If orbit belongs to some curve $f(x, y) = \rm{const}$, then at least you could obtain by differentiantion that $(\nabla f, X) = 0$ along this curve. It's easy to show that by chain rule: suppose that $(x(t), y(t))$ is a trajectory that belongs to curve $f(x, y)= \rm const$, then $f(x(t), y(t)) = \rm{const}$ and $\frac{\partial f}{\partial x} \cdot x'_t + \frac{\partial f}{\partial y} \cdot y'_t = 0$ which can be rewritten as $(\nabla f, X)$.