Suppose that $f(x)$ is an irreducible cubic polynomial with integral coefficients. Suppose further that for all primes $p$, there exists an integer $n_p$ for which $p^2 \nmid f(n_p)$. Then it is a theorem of Erdős that $f(x)$ takes on infinitely many square-free values. This result was refined by Hooley, and extended to the situation where $f(p)$ takes on infinitely many square-free numbers where $p$ ranges over the rational primes by Helfgott in 2013, which was essentially proved in a different way by Reuss in 2015.

Consider, for a positive parameter $B$, the set $$\displaystyle S_f(B) = \{t \in \mathbb{N} : t \leq B, t \text{ is square-free}, \exists x \in \mathbb{Z} \text{ s.t. } f(x) = t\}.$$

I want to know the relative density of $S_f(B)$ in the set

$$\displaystyle T_f(B) = \{t \in \mathbb{N}: t \leq B, \exists x \in \mathbb{Z} \text{ s.t. } f(x) = t\},$$

that is, the quantity

$$\displaystyle \lim_{B \rightarrow \infty} \frac{\# S_f(B)}{\# T_f(B)}$$

if the limit exists.

It seems that the relative density should be higher than $6/\pi^2$, since there is a bias towards certain primes. In particular, if $t$ is a number which is divisible by a prime $p$ for which $f(x) \equiv 0 \pmod{p}$ has no solution, then $t$ cannot be in either set. Therefore, I suspect Chebotarev's density theorem to enter the picture somehow.

Any insight would be appreciated.