Here is a problem I thought I solved but now I think it can't be right.
The problem is as follows: Let $X$ be a scheme and $\mathcal{F}$ a coherent sheaf such that every $x\in X$ has an étale neighborhood $f:U\to X$ such that $f^*\mathcal{F}$ is free, then $\mathcal{F}$ is locally free on $X$.
Here is my attempt: Let $x\in X$ be any point, $f:U\to X$ an étale neighborhood, $y\in U$ and $f(y)=x$. In particular $f$ is flat so $\mathcal{O}_x\to \mathcal{O}_y$ is flat. Since it is a local homomorphism of rings it is in fact faithfully flat (Matsumura, corollary to 4A). Now $\mathcal{F}_x\otimes_{\mathcal{O}_x}\mathcal{O}_y=(f^*\mathcal{F})_y$ is free by assumption. It then follows that $\mathcal{F}_x$ is free $\mathcal{O}_x$ module (Matsumura 4E).
This approach is not really good because I need $X$ noetherian to conclude that $\mathcal{F}$ is locally free. Also, I'm not sure if the argument is even correct because only flatness of $U\to X$ is being used. If anyone can point out an error in my solution and/or give me a hint on how to attack this problem I would be grateful!
Best Answer
First, Hartshorne only defines an etale morphism for schemes of finite type over a field (exercise III.10.3), so if you want to assume $X$ is noetherian and use Hartshorne's definition of a coherent sheaf as a quasi-coherent sheaf of finite type, you can proceed exactly as in your post and you really do only use flatness. If you upgrade your definition of coherent sheaf to be one which is affine-locally the sheaf associated to a coherent module as Aphelli does in the comments, which is equivalent to the definition Hartshorne uses in the noetherian case, that proof will work as well.
Indeed, the full strength of the map being etale is not necessary in general: for quasi-coherent sheaves, being finite locally free is fpqc local (ref Stacks 05B2, for instance). The proof is that flat + finite presentation = locally free of finite rank, and both of the first two ingredients are fpqc-local.