Yes. At least in general. In fact, answering whether any SAT (or 3SAT) instances are hard equivalent to settling P vs. NP. Assuming easy = polynomial, the "easy subset" of SAT = SAT iff P = NP. Separating hard from easy instances -- in general -- is pretty much the secret to complexity theory.
We do know to recognize some easy cases. SAT is FPT -- i.e. fixed parameter tractable [1] -- because its worst-case solving time is exponential in the number of variables. There are more complex cases of easy instances.
A question that is easier than P vs. NP is separating SAT instances that are easy/hard for DPLL (+ improvements), the 1960s algorithm at the core of all SAT solvers. AFAIK, this, too, is an unanswered question in general (i.e., given a SAT instance, we don't know how to tractably determine whether or not DPLL will tractably solve it).
[1]: https://en.wikipedia.org/wiki/Parameterized_complexity#FPT
Usually, the practice for solving these sorts of problems is to have your algorithm return three values: the solution, unsatisfiable, or timed out. There is literature on how to structure such problems to make it more likely you're going to get answers, but I'm not personally familiar with this area.
There is no such proof for the subset of 3SAT that we can solve quickly. It's for the entire problem only.