Collisionless Boltzmann Equation

I think it basically boils down to the fact that the gradient of the gravitational potential energy is the relevant force for your problem. For other problems, this needn't be true. Maybe you're dealing with charged particles so you're also interested in the electric potential energy.

I'm going to rewrite things in a different notation since I think it'll make things clearer.

Your distribution function $f$ is a function of position $vec{r}$, velocity $vec{v}$ and time $t$. Assuming there are no collisions

$fleft(vec{r}, vec{v}, tright) = fleft(vec{r} - vec{v}dt, vec{v} - frac{vec{F}}{m}dt, tright),$

where $vec{F}$ is the force that accelerates your particles. Taking the derivitives

$frac{df}{dt} = frac{partial f}{partial vec{r}} cdot vec{v} + frac{partial f}{partial vec{v}} cdot frac{vec{F}}{m} + frac{partial f}{partial t}.$

Generally speaking, $vec{F} = frac{partial U}{partial vec{r}}$, where $U$ is the potential energy. Since you're dealing with the gravitational potential (not the gravitational potential energy), you have that $frac{vec{F}}{m} = frac{partial Phi}{partial vec{r}}$.

So in your notation, $frac{partial vec{v}}{partial t}$ = acceleration = $frac{vec{F}}{m} = frac{partial Phi}{partial vec{r}}$.

For any conservative force one can define the field, $mathbf{A}$, associated with the force as the gradient of a scalar potential, $phi$, or: $$ mathbf{A} = -nabla phi tag{0} $$

We know from Newton's laws that a force is just the mass times acceleration and that $mathbf{a} = dmathbf{v}/dt$. Thus, in your example the gravitational potential is given by: $$ frac{dv_{i}}{dt} = a_{i} = frac{F_{i,grav}}{m} = -nabla_{i} Phi_{grav} = -frac{d Phi_{grav}}{dx_{i}} tag{1} $$