CNF-SAT reduction problem variant
1/3-CNFSAT is very much similar to HALF-SAT, where instead of satisfying exactly half of the total number of clauses you are satisfying exactly one third of it. So the reduction from 3SAT is also quite similar. We reduce 3SAT instance, $\phi$, to a 1/3-CNFSAT instance as follows.

Let $z$ be a new variable not in $\phi$. Let there be $m$ clauses in $\phi$. We add $(z \lor \lnot z)$ and $2m+2$ copies of $(z)$ to $\phi$. In every exactly 1/3-satisfying assignment $z$ must be false.

Let $\phi \land (z \lor \lnot z) \land (z) \land (z) \land ... (2m+2) $ copies $...\land (z) = \phi'$. The following claim is easy to prove.

Claim: $\phi$ is satisfiable iff $\phi'$ has exactly 1/3rd satisfiable clauses.

Therfore 3SAT $\leq_P$ 1/3-CNFSAT, and hence 1/3-CNFSAT is NP-complete.