Non conformally geodesible vector field What is  an example  of  a  smooth  vector  field  $V$ on an open set of  the  plane which is  a  geodesible  vector  field but  there is  no  a  conformal  metric $g$ such that $V$ is  geodesible  vector  field  with respect to $g$.
A geodesible  vector  field is a non vanishing vector  field for which there is  a  Riemannian metric $g$ such that all trajectories of $V$ are (unparametrized) geodesics with respect to $g$.
The  motivation for  this  question is  described here:
Limit cycles of quadratic systems and closed geodesics
 A: Here is how one can construct an example: Consider the smooth, nonvanishing $1$-form 
$$
\omega =  y^3(1{-}y)^2\,\mathrm{d}x + \big(y^3-2(1{-}y)^2\bigr)\,\mathrm{d}y.
$$ 
Note: This $\omega$ came from Exercises 5 and 6 of Section 16 of Chapter XVIII of Volume IV of Dieudonné's Treatise on Analysis.  These exercises show that $\omega$ cannot be written globally in the form $g\,\mathrm{d}f$ for two smooth functions on $\mathbb{R}^2$.  (In other words, $\omega$ has no global 'integrating factor'.)
I am going to use Dieudonné's $1$-form to show that the nonvanishing vector field
$$
V = y^3(1{-}y)^2\,\frac{\partial\ }{\partial x} 
+ \big(y^3-2(1{-}y)^2\bigr)\,\frac{\partial\ }{\partial y},
$$
while geodesible, is not geodesible with respect to any conformal metric on the plane, i.e., a metric of the form $g = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$.
First, I will show that $V$ is geodesible.  To do this, it suffices to find a closed $1$-form $\phi$ such that $\phi(V)>0$.  I construct $\phi$ as follows:  Let $\rho \approx 0.639$ be the unique real root of $\rho^3-2(1{-}\rho)^2 = 0$.  Now set
$$
f(y) = \frac{\rho^3(1{-}\rho)^2-y^3(1{-}y)^2}{y^3-2(1{-}y)^2},
$$
and note that $f(y)$ is a rational function of $y$ with a quadratic denominator that never vanishes.  Hence $f(y)$ is a smooth function of $y$.  Now set
$$
\phi = \mathrm{d}x + f(y)\,\mathrm{d}y.
$$
Then $\phi$ is closed and computation yields $\phi(V) = \rho^3(1{-}\rho)^2>0$.  Thus, $V$ is geodesible.
To see that $V$ not geodesible with respect to any metric of the form $g= \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$, write
$$
V = h(y)\left(\cos\theta(y)\,\frac{\partial\ }{\partial x} 
+ \sin\theta(y)\,\frac{\partial\ }{\partial y}\right)
$$
where $h(y)>0$, and consider, for any function $u(x,y)$, the two $1$-forms
$$
\eta_1 = \mathrm{e}^{u(x,y)}\bigl(\cos\theta(y)\,\mathrm{d}x{+}\sin\theta(y)\,\mathrm{d}y\bigr)
\ \ \text{and}\ \ 
\eta_2 = \mathrm{e}^{u(x,y)}\bigl(-\sin\theta(y)\,\mathrm{d}x{+}\cos\theta(y)\,\mathrm{d}y\bigr).
$$
If $V$ is to be geodesic with respect to the conformal metric
$$
{\eta_1}^2+{\eta_2}^2 = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2),
$$
then one must have $\mathrm{d}\eta_1 = 0$, which would imply, since $\mathbb{R}^2$ is simply connected, that $\eta_1 = \mathrm{d}v$ for some function $v=v(x,y)$ on $\mathbb{R}^2$.  However, by construction,
$$
\mathrm{d}v = \eta_1 = \mathrm{e}^{u(x,y)}\,\frac{\omega}{h(y)},
$$
so $\omega = \mathrm{e}^{-u(x,y)}h(y)\,\mathrm{d}v$, which is impossible by Dieudonné's Exercises.  Thus, $V$ is not geodesible with respect to any conformal metric on $\mathbb{R}^2$.
