I think it is something like a folkore result that a coherent sheaf $\mathcal F$ on a smooth algebraic variety $X$ over $k$, which is equipped with a connection
$\nabla: \mathcal F \rightarrow \mathcal F \otimes \Omega^1_{X/k}$
is already locally free.
Maybe one may weaken the assumptions, but I think the proof wouldn't alter very much.
I thought about how to prove it, but couldn't make it rigorous. In any case one should show that the stalks $\mathcal F_x$ are free, and this somehow must follow from the existence of the connection.
Addendum: It seems that the existence of a connection in some way rigidifies the underlying sheaf.
A related question in this respect is: in how much is a horizontal morphism $\phi: \mathcal F \rightarrow \mathcal F$ already rigidified by $\nabla$. E.g. if one knows that $\phi$ is the identity on $\mathcal F(x) \rightarrow \mathcal F(x)$, then is it so already around $x$?
Best Answer
See Prop. 8.8, p. 206 in N. Katz, "Nilpotent connections and the monodromy...", Publications Mathématiques de l'IHES, 39 (1970), p. 175-232.