Consider two planar vector fields and the height of their cross product: \begin{align} V&=P(x,y)\,\partial_x+Q(x,y)\,\partial_y\\ W&=\ \ \ \ \ x\partial_x\ \ \ \ +\ \ \ \ \ y\partial_y\\ f(x,y)&= P(x,y)\,y\ -\,Q(x,y)\,x\\ \end{align} where $P,Q \in \mathbb{R}[x,y]$ are polynomials of degree $2$ with $P(0,0)=Q(0,0)=0$.

These vector fields lead to a Riemannian metric on $\mathbb{R}^2 \setminus f^{-1}(0)$ whose orthonormal frame is: $$\left\{\frac{x^2+y^2}{f(x,y)}V,\ \frac{1}{x^2+y^2}W\right\}.$$

Call the curvature of this metric $\kappa$.

Question: When a quadratic vector field $V$ does not have a center on the plane, is the curve $$\{(x,y)\mid \kappa(x,y)=0\}$$ transverse to $V$?

If not, what rescaling $g(x,y)W$ in the orthonormal frame would give a positive answer?

Notes:

A center is a singularity which is surrounded by a band of closed orbits. For quadratic vector fields they are classified at this paper.

The trajectories of $V$ are the geodesics of this metric. On a periodic orbit of $V$ which surrounds the origin, $\,f$ is never 0.

The motivation for this post is mentioned in this answer and this post.