That set is the locus of function ##f##, and is a subset of ##\Omega\times \mathbb R##, which in turn is a subset of ##\mathbb R^{m+1}##.
By analogy with the locii of function from ##\mathbb R\to\mathbb R## ( a line, ie one-dimensional manifold in a 2D space), and from ##\mathbb R^2\to \mathbb R## (a surface or 2D manifold in a 3D space), we'd expect the set to be a ##m##-dimensional manifold.
Your mission, should you choose to accept it, is to, for any arbitrary point in the set, construct a homeomorphism from an open neighbourhood of the point to an open subset of ##\mathbb R^m##. If you can do that, you've proven the set is a ##m##-dimensional manifold.
Construction of that homeomorphism may involve the function ##f## in some way, or at least use its property of smoothness.
EDIT: Just realised the proposition is not true unless we require ##\Omega## to be a manifold. Consider where ##\Omega = \{0\} \cup [1,2]## and ##f(x)=x##. Then neither the domain nor the locus of ##f## is a manifold. They are each the union of a 0-dimensional manifold with a 1-dimensional manifold with boundary.