r-regular graph and hamiltonian path
You are underselling the "G or G*" part. This suggests a proof of the form:


"assume G isn't Hamiltonian ... prove G* must be". So try that route.


To give a further hint, I remind you of the Dirac theorem:


If a connected graph G has n >= 3 vertices and the degree of each vertex is at least n/2, then G has a Hamiltonian cycle (if it has a cycle, obviously it also has a path, you just remove an edge from the cycle).
Additionally, an n-vertex r-regular graph with r >= n/2 must be connected (assume it is not connected - what is the minimum number of vertices each connected component must have? Can there be more than 1 such component?)


I think these should give you the idea how to prove it for n >= 3.

Then it just remains to prove it for n = 2, which is trivial (the only reason it isn't included in above theorem is that it only has a path, not a cycle).

Note: because I am not sure a complement is even defined on a non-simple graph, I am assuming G is simple. But if it wasn't, the statement-to-prove couldn't be true, even if you defined a complement in any way that makes sense - what would be the complement of this graph, and how could either have a Hamiltonian path: