If $U$ is a function of $x,y,z$ then the total increment of $U$ due to a transition of a point from $(x,y,z)$ to $(x+\delta x,y+\delta y,z+\delta z)$ is given by the total differential given by
$$dU = \frac {\partial U}{\partial x} dx + \frac {\partial U}{\partial y} dy +\frac {\partial U}{\partial z} dz$$
which implies
$$dU = ( \frac {\partial U}{\partial x} \hat i + \frac {\partial U}{\partial y} \hat j +\frac {\partial U}{\partial z} \hat k) \cdot (dx \hat i+dy \hat j+dz \hat k)$$
$$dU = \nabla U \cdot dl$$
where $$\nabla U = \frac {\partial U}{\partial x} \hat i + \frac {\partial U}{\partial y} \hat j +\frac {\partial U}{\partial z} \hat k$$ and $$dl= (dx \hat i+dy \hat j+dz \hat k)$$
but if the total differential is zero what does it say about the gradient other than the fact that this implies that there is no change in U.
Edit:
$dU=0$ does not necesaarily mean change in U is zero but is or order higher than 2
It does NOT imply that there is no change in $U$.
What it implies is that the change in $U$ is of order 2 or higher; or to put it another way, the best "polynomial" approximation is a polynomial of degree 2 or higher. Think for example of $f(x,y,z)=x^2+y^2+z^2$ at $(x,y,z)=(0,0,0)$.