Check this link for the question: total differential means exact in this case
The question want to show $ I\eta$ is not exact ($I\eta \neq dq$, where d is exterior derivative and q is 0-form). I am failing to prove that $I\eta$ is not $C^{1},C^{2},C^{3}$ otherwise $I\eta$ is exact. Any help would greatly appreciated.
SKETCH: I will switch to the letter $f$ for the function. Suppose $f\eta$ is exact. If so, then it must be closed, so $$d(f\eta) = df\wedge\eta + f\,d\eta = 0.$$ This implies that $$d\eta\wedge\eta = 0.$$ (Why?) Does this equation hold?