In spherical coordinates, the metric of $\mathbb{R}^3$ (Euclidean space) has the following nice properties:
- $g_{rr}$ is independent of $r$, $\theta$ and $\phi$
- $g_{\theta\theta}$ is independent of $\theta$ and $\phi$
- $g_{\phi\phi}$ is independent of $\phi$
- $g_{r\theta} = g_{r\phi} = g_{\theta\phi} = 0$
That's 3 components of the metric, and 6 partial derivatives, that are zero. Pretty convenient to work with!
So what about a general Riemannian 3-manifold that is diffeomorphic to $\mathbb{R}^3$ but has curvature? How many of those 9 conditions can we enforce simultaneously, by choosing just the right "spherical" coordinate system?
For example, once we pick an origin, can we claim that the geodesics emanating from the origin will never intersect each other? In that case, we could make them our $r$ coordinate lines, have $g_{rr}$ measure true distance, and thus satisfy the first point: $g_{rr,r} = g_{rr,\theta} = g_{rr,\phi} = 0$. And wouldn't that also imply $g_{r\theta} = g_{r\phi} = 0$?
And is it possible to do this in such a way that at least one of $g_{\theta\phi}$, $g_{\theta\theta,\theta}$, $g_{\theta\theta,\phi}$, and $g_{\phi\phi,\phi}$ also vanish globally? How many could we get?
Are you asking about intrinsic properties of a given metric (regarding this: your example with spherical coordinates on $\mathbb R^3$ does NOT define a curved metric) or about properties of specific coordinate systems? For standard metric on $\mathbb R^3$ you can of course have all components of metric constant, just use cartesian coordinates. For more general metrics this is usually not possible because there is an obstruction given by the curvature tensor (if metric components are constant in some coordinate system, then curvature tensor vanishes, but of course there exist metrics with nonzero curvature tensor).