$Xf=0$ on $\{f=0\}$ implies $f$ divides $Xf$?

Question

Suppose $f\in \mathbb{R}[x,y]$ is irreducible such that $\nabla.f=(f_x,f_y)\neq 0$ on $\{f=0\}$. So, $\{f=0\}$ is a regular curve. Suppose $X=(P,Q)$ is a vector field on $\{f=0\}$, i.e., $Xf=Pf_x+Qf_y=0$ on $\{f=0\}$ where $P,Q\in \mathbb{R}[x,y]$. Is it true that $f$ divides $Xf$.

This is of course true in any algebraically closed field due to Hilbert Nullstellensatz. But I tried to construct counter examples on $\mathbb{R}$ but I failed.

Answer 1

Let $f=x^2+y^2+1$ and $X=(1,0)$, so $Xf=2x$. Then $f$ does not divide $Xf$.