How is this reduction of 3-SAT to Half-SAT not valid? I am studying algorithms and there is a question in CLRS called the Half-SAT problem

We are given a 3-CNF formula with n variables and m clauses where m is even. We wish to determine whether there exists a truth assignment to the variables of f such that exactly half the clauses evaluate to zero and the other half to 1. Prove that the half 3-SAT problem is NP-Complete.

What is to stop me from taking any input f to 3-SAT and adding m clauses (y or y or y) where y is not in the original m clauses. The transformation of input and output is clearly polynomial time.
And, given a solution to 3-SAT, there must be a satisfying solution to Half-SAT (the solver sets y to false). If there is not a solution to 3-SAT this means that one of the original m clauses is not satisfied these clauses are disjoint from the new clause, so either m - 1 or 2m - 1 clauses are True.
Yet all the solutions online are more complicated which makes me think I am missing something obvious.
As mentioned in the comments, this reduction is incorrect because it is necessary to have equivalence. While we may have 3-SAT =&gt; Half-SAT we do not have !3-SAT =&gt; !Half-SAT or Half-SAT =&gt; 3-SAT (the contrapositive). A proper reduction would need to do something like so:
Add m + 1 clauses (y or y or y)
Add a contradictory clause (y or not y)
Now exactly half of the clauses are True iff there is a solution to 3-SAT.