This example is confusing me.
Is $\mathbb{Z}_3 = \langle a\vert a^3\rangle$ $\operatorname{CAT}(0)$ and/or (Gromov) $\delta$-hyperbolic?
The Cayley graph clearly has bounded diameter, therfore it is (Gromov) $\delta$-hyperbolic.
But triangles are more "fat" than comparison triangles in $\mathbb{R}^2$. Which would mean that it is not CAT(0).
I also read that it is an open question if all Gromov hyperbolic groups are also CAT(0). Therefore for this simple example I think it is more likely that the group is in fact CAT(0).
I can't see where I make a mistake.
Edit: As pointed out by Lee Mosher in comments, it is sufficient to say that finite groups act geometrically on a point: of course, a point is hyperbolic and CAT(0)...
Finite groups are clearly hyperbolic, therefore $\mathbb{Z}_3$ so is.
Then, whether a group is CAT(0) is not readable on its Cayley graph: we want to know if it acts geometrically on a CAT(0) space. In the case of $\mathbb{Z}_3$, you can make it act on a disk $D^2$ thanks to a rotation of order 3: $D^2$ is CAT(0) and the action is geometric, so $\mathbb{Z}_3$ is a CAT(0) group.
More generally, it is not difficult to prove that any finite group $F$ is CAT(0): let $T$ be the graph whose vertices are $F \cup \{ o \}$ and where $o$ is linked by an edge with any vertex of $F$ - therefore, $T$ looks like a star. Now, let $F$ act on $T$ fixing the vertex $o$. Clearly, $T$ is CAT(0) - it is a tree - and the action $F \curvearrowright T$ is geometric.