Completeness of solution to a differential equation for a vector field In the context of a research problem on the QM magnetic response of electrons, I came across a differential equation:
$$ \vec{v} \times \vec{\nabla} \ln{\rho}  = \vec{\nabla} \times \vec{v} \tag{1}$$
for a vector field $\vec{v}: \mathbb{R}^3 \to \mathbb{R}^3$ and (very)well-behaved (continuous, differentiable, and bounded & quickly converging to $0^+$ for large arguments) $\rho: \mathbb{R}^3 \to (0,1)$. Solutions for $\vec{v}$ in dependence of $\rho$ are sought.
My attempts have led me to the conclusion that all $$\vec{v}(\vec{x})=\vec{\nabla}G(\rho(\vec{x})) \tag{2}$$ solve the equation for arbitrary real functions $$G: \mathbb{R} \to \mathbb{R}$$
Initially I thought that this would characterize the complete set of solutions in the sense that functions of any other form than $(2)$ violate $(1)$. After some numerical experiments it appears to me that I am missing other solutions.
Can anyone comment on the completeness of $(2)$? A typical example for $\rho$ would be
$\rho(\vec{r}) = \sum_{i=0}^{n} a_i \exp{(b_i(\vec{x}-\vec{x}_i)^2)}$ with $a_i,b_i \in \mathbb{R}^+$ and small $n$ like $2$.
Are there counterexamples to these solutions? What would be a more general solution?

Edit
$(1)$ can be rewritten to
$$ \rho\vec{\nabla}\times\vec{v} = \vec{\nabla}\rho\times\vec{v} \tag{3}$$
possibly this is helpful.
 A: If we write out the equations explicitly (letting $\ln \rho = q$ for convenience) we have
\begin{align}
\partial_{y} v_{3} - \partial_{z} v_{2} &= v_{2} \partial_{z} q - v_{3} \partial_{y} q \implies \partial_{y} v_{3} + v_{3} \partial_{y} q = \partial_{z} v_{2} + v_{2} \partial_{z} q \\
\partial_{z} v_{1} - \partial_{x} v_{3} &= v_{3} \partial_{x} q - v_{1} \partial_{z} q \implies \partial_{z} v_{1} + v_{1} \partial_{z} q = \partial_{x} v_{3} + v_{3} \partial_{x} q \\
\partial_{x} v_{2} - \partial_{y} v_{1} &= v_{1} \partial_{y} q - v_{2} \partial_{x} q \implies \partial_{x} v_{2} + v_{2} \partial_{x} q = \partial_{y} v_{1} + v_{1} \partial_{y} q \\
\end{align}
Multiply each equation by $e^{q}$ and reverse the product rule
\begin{align}
\partial_{y} (e^{q} v_{3}) &= \partial_{z} (e^{q} v_{2}) \\
\partial_{z} (e^{q} v_{1}) &= \partial_{x} (e^{q} v_{3}) \\
\partial_{x} (e^{q} v_{2}) &= \partial_{y} (e^{q} v_{1}) \\
\end{align}
which is to say
\begin{equation}
\nabla \times (e^{q} v) = \vec{0}
\end{equation}
i.e $e^{q} v$ is irrotational.
Assuming the domain is simply connected, the general solution is then given by
\begin{equation}
e^{q} v = \nabla \phi \implies v = e^{-q} \nabla \phi = \rho^{-1} \nabla \phi
\end{equation}
for an arbitrary differentiable function $\phi$.
Edit
We can see that this works by substituting the result into the original equation
\begin{align} \nabla \times v &= \nabla \times (\rho^{-1} \nabla \phi) \\ &= \rho^{-1} \nabla \times (\nabla \phi) + \nabla (\rho^{-1}) \times \nabla \phi \\ &= (- \rho^{-2} \nabla \rho) \times \nabla \phi \\ &= \rho^{-2} \nabla \phi \times \nabla \rho \\ &= (\rho^{-1} \nabla \phi) \times (\rho^{-1} \nabla \rho) \\
&= v \times \nabla \ln \rho \end{align}
