I was asking myself this question. We define for an exact vector field $\mathbf{v}(\mathbf x) \in \mathbb{R}^n$, s.t. $\mathbf{v}(\mathbf x)=\nabla u(\mathbf x)$ for a potential $u$, a solution as a function $f(t):[0,T] \rightarrow \mathbb{R}^n$ such that $f'(t)=\mathbf{v}(f(t))$ for every $t$.
The question is: if the vector field is exact, can there be solutions with $f(0)=f(T)$, i.e. closed loops ?
Observation: If we have the hypothesis that the vector field is always non zero maybe the situation is easier. In fact if such a solution would exist and we call $\gamma$ the associated loop, $\int_{\gamma} \mathbf{v}=\int_{\gamma} \nabla{u}=\mathbf{0}$, so the integral of the vector field along the loop should be zero. This means that the component tangent to the loop should change sign, in particular there must be a point $p$ along the loop where the component of the vector field tangent to the loop vanishes. This should be true for every loop. For our loop the vector field is also always tangent to $\gamma$, so this means that $\mathbf{v}(p)=\mathbf{0}$. But this cannot be if the vector field is always non zero.
Is the observation correct or formalizable with enough regularity conditions? And what about exact vector fields that can vanish? Are there general results on the loops that an exact vector field can have?
First of all notice that if $\mathbf{v}$ has a zero, say in $\mathbf{x}_0$, then a solution trivially exists by choosing $f(t)\equiv\mathbf{x}_0$.
Your observation is correct. Indeed, we find for any solution $f\in C^1([0,T];\mathbb{R}^n)$ by the chain rule $$0=\int_0^T[u(f(t))]'\mathrm{d}t=\int_0^T\nabla u(f(t))\cdot f'(t)\mathrm{d}t=\int_0^T\mathbf{v}(f(t))\cdot f'(t)\mathrm{d}t=\int_0^T\vert f'(t)\vert^2\mathrm{d}t$$ and thus $f\equiv f_0\in\mathbb{R}^n$ must be constant. From this we conclude $$\mathbf{v}(f_0)=f_0'=0.$$