Is the curvature of space-time a smooth function everywhere (except at black holes) in view of General relativity. By 'smooth' it is meant that it possesses derivatives of all order at a given point.
No, not at the boundary of a solid object like a planet. There's a step function in the stres-energy tensor, and so you'll have a step function in the Riemann tensor.
I actually believe the answer to this question is yes, and the reason is that a fundamental assumption of GR is that the gravitational field is constant over small regions of space time. Another way of phrasing this is to say the GR is only valid in the limit of geometrical optics, which is the limit for which the wavelength of light is small compared with the length scale of the problem under consideration.
It's very hard to do any calculations in GR, and the 2 well known cases (black holes and cosmology) both involve smooth functions for all intents and purposes. Actually I think it's generally accepted that almost all functions that arise in physics are analytic (i.e. can be represented by power series), which except for some pathological examples means the functions are smooth.
Well other than black hole singularities, there are other cosmological topological defects like cosmic strings, domains, and textures (and magnetic monopoles) that would not be considered as smooth. These are regions where there is or has been a phase transition. Of course, these things may not exist, but string theory predicts monopoles, so they had better ;)