"When $f(x)$ is prime, it is a Cuban prime."
This doesn't seem to be true. For example, take $x=2$. Then $f(x)= \lfloor 3 \rfloor=3$, which is prime. On the other hand, $3$ is not a Cuban prime, as the smallest Cuban prime is $7$.
It appears, by your references, that you are instead considering generalized Cuban primes. That is, whereas Cuban primes are simply those arising as the difference of successive cubes, generalized Cuban primes take the form
i.e. as a difference of powers, divided by one trivial factor (without this division, we would find no generalized Cuban primes that are not Cuban themselves).
Now turning to your problem, suppose that $f(x)$ is prime. If $x=2k$ is even, then
$$f(2k)= \lfloor 3(2k)/2 \rfloor = \lfloor 3k \rfloor = 3k,$$
which is composite unless $k=1$ (i.e. $x=2$ and $f(x)=3$). Otherwise, suppose that $x=2k+1$ is odd. Then
$$f(2k+1)= \lfloor (3(2k+1)/2\rfloor=\lfloor 3k+3/2 \rfloor =3k+1.$$
Thus $f(x)$ prime implies $p \equiv 0,1 \!\!\mod 3$, so that prime values of $f$ are in fact (generalized) Cuban primes. Of course, since $f$ surjects onto the set of integers congruent to $1$ modulo $3$, the two sets you consider are in fact equal.
Why, then, is your definition non-standard?
- The floor function implicitly introduces cases, in situations that may not require casework.
- The condition $p \equiv 0,1 \!\!\! \mod 3$ makes the properties of $p$ very clear.
- The other two definitions for (generalized) Cuban primes are given in terms of binary quadratic forms. These have widespread use in number theory, and it is useful (yet sometimes non-obvious) to know when a set is related to such functions.
In short, while your definition is elegant (but perhaps less so than $p \equiv 0,1 \!\!\mod 3$), it is far from transparent, which is often more useful in mathematics.