Trivially
$$
f_n(q)=\frac{\left(q^3;q^3\right)^4_{\infty }}{(q;q)_{\infty } \left(q^9;q^9\right){}_{\infty }}.
$$
Denoting $A(q)=(q;q)_{\infty }(q^2;q^2)_{\infty }$, one can see that in order to prove OP's claim it is enough to perform trisection of $A(q)$, which can be found in Michael Somos' article [A Multisection of q-Series ][1]. Using formulas from section 4 of this article, by simple algebra one can find
$$
A(q \xi)\xi^2+{A\left({q}{\xi}^2\right)}{\xi}=-A(q)-3qA(q^9).
$$
This reduces OP's claim to verification of an eta-product identity:
$$
A\left(q^3\right)^4-\left(3 q A\left(q^9\right)+A(q)\right) A\left(q^9\right) A(q)^2-9 q^2 A\left(q^9\right)^3 A(q)=0.
$$
For this purpose, one can consult Somos' [table][2] of such identities. With high probability it can be found there. Even if it can not be found, identities for eta-products can be verified automatically using modular machine.

Thus the identity is related to Ramanujan's famous cubic continued fraction, in other words very classical, textbook stuff.


  [1]: https://grail.eecs.csuohio.edu/~somos/multiq.html
  [2]: https://web.archive.org/web/20190709150645/https://eta.math.georgetown.edu/index.html