It is quite likely that the original cubic polynomial is the same, under a linear invertible change of variables with rational integer coefficients, to the one in the Corollary on the final page of Nowlan 1926.
The behavior as far as represented primes $\pm 1 \pmod 9$ and primes giving only trivial zeroes $2,4,5,7 \pmod 9$ (as your 13: i am calling this "anisotropic") is identical, and the polynomials are quite similar. There are people on this site who could determine this by using the number field in question, I imagine Jyrki could do that, so I left him an additional note. For me, I am doing some trial and error.
EEEEDDDDIIITTT: It worked. The polynomial given by the OP is equivalent, by an invertible linear change of variables with rational integer coefficients, to Nowlan's 1926 polynomial.
As a result, we have Nowlan's Corollary on the final page: for (positive) primes $p \equiv 2,4,5,7 \pmod 9,$ the OP's polynomial can only be divisible by $p$ if all three of his $a,b,c$ are divisible by $p.$ Meanwhile, for prime $q=3$ or $q \equiv \pm 1 \pmod 9,$ we can find rational integers $a,b,c$ such that $f(a,b,c) = p.$ Finally, if the polynomial integrally represents two numbers, it represents their product. So, we can tell exactly what numbers are integrally represented by the polynomial: for every prime factor $p \equiv 2,4,5,7 \pmod 9,$ the exponent must be divisible by $3.$ Those are the only restrictions.
Meanwhile given Nowlan's
$$ g(u,v,w) = u^3 + 6 u^2 w - 3 u v^2 + 3 uvw +9 u w^2 - v^3 + 3 v w^2 + w^3, $$
$$ g(a+c, -a-b, -a) = a^3 + b^3 + c^3 -3abc -3 a^2 b - 3 b^2 c - 3 c^2 a = f(a,b,c) $$
We can invert this with
$$ f(-w, w-v,u+v) = g(u,v,w). $$
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parisize = 4000000, primelimit = 500509
? u = a+c
%1 = a + c
? v = -a - b
%2 = -a - b
? w = -a
%3 = -a
? p = u^3 - v^3 + w^3 + 3 * u * v * w - 3 * u * v^2 + 3 * v * w^2 + 9 * u * w^2 + 6 * u^2 * w
%4 = a^3 - 3*b*a^2 + (-3*c^2 - 3*b*c)*a + (c^3 - 3*b^2*c + b^3)