Trivially $$ f_n(q)=\frac{\left(q^3;q^3\right)^4_{\infty }}{(q;q)_{\infty } \left(q^9;q^9\right){}_{\infty }}. $$$$ \prod_{n\geq1} f_n(q)=\frac{\left(q^3;q^3\right)^4_{\infty }}{(q;q)_{\infty } \left(q^9;q^9\right){}_{\infty }}. $$ 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 . 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 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.