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.](https://www.researchgate.net/publication/246484433_Integrability_of_plane_quadratic_vector_fields)

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](https://mathoverflow.net/questions/273635/finding-a-1-form-adapted-to-a-smooth-flow/273648#273648) and [this post.](https://mathoverflow.net/questions/160945/limit-cycles-as-closed-geodesicsin-negatively-or-positively-curved-space)