Let $u :\mathbb R^d \to \mathbb R$ be smooth and $v := \nabla u$ its gradient. We fix $x_0$ and let $x_t$ be the solution of $\dot x_t = v(x_t)$ started from $x_0$. Let $\rho_t := \delta_{x_t}$ for all $t \ge 0$. Then @leo said in his comment that
$\rho_t$ solves the continuity equation $\partial_t\rho_t+\nabla\cdot(\rho_t v)$ in the sense of distribution, i.e., $$ \int_0^\infty \int_{\mathbb R^d} \left(\partial_t \phi(x, t)+\left\langle v(x), \nabla_x \phi(x, t)\right\rangle\right) \mathrm d \rho_t(x) \mathrm d t=0 \quad \forall \phi \in \mathcal C^\infty_c(\mathbb R^d \times (0, \infty)). $$
My understanding It follows from $\rho_t := \delta_{x_t}$ that $$ \begin{align} &\int_{\mathbb R^d} \left(\partial_t \phi(x, t)+\left\langle v(x), \nabla_x \phi(x, t)\right\rangle\right) \mathrm d \rho_t(x) \\ ={} &\partial_t \phi(x_t, t)+\left\langle v(x_t), \nabla_x \phi(x_t, t)\right\rangle \\ ={} &\partial_t \phi(x_t, t)+\left\langle \dot x_t, \nabla_x \phi(x_t, t)\right\rangle. \end{align} $$
Then we need to show that $$ \int_0^\infty (\partial_t \phi(x_t, t)+\left\langle \dot x_t, \nabla_x \phi(x_t, t)\right\rangle) \mathrm d t=0. $$
Could you elaborate on how to finish the proof?