It turns out that Ramanujan has asserted a result, from which the identity in the question follows. In order to state his result, recall the $q$-Pochhammer symbols $(a;q)_\infty = \prod_{k \ge 0}(1-aq^n)$. For convenience, we always suppress the subscript $(a;q) = (a;q)_\infty$.
Ramanujan asserted that,
$$(q;q) = \frac{(q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25})}{(q^{5};q^{25})(q^{20};q^{25})} - q(q^{25};q^{25})-q^2\frac{(q^{5};q^{25})(q^{20};q^{25})(q^{25};q^{25})}{(q^{10};q^{25})(q^{15};q^{25})}.$$
The pentagonal number theorem says that the left hand side $(q;q) = \sum_n (-1)^nq^{G_n}$. However the right hand side decomposes the sum based on $G_n\bmod 5$. Using the notation $F_k$ defined in the question, we obtain $$F_0 = \frac{(q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25})}{(q^{5};q^{25})(q^{20};q^{25})}, F_1 = - q(q^{25};q^{25}), F_2=-q^2\frac{(q^{5};q^{25})(q^{20};q^{25})(q^{25};q^{25})}{(q^{10};q^{25})(q^{15};q^{25})}.$$
Clearly, $F_0F_2 + F_1^2 = 0$.
Inspired by the proof of the Ramanujan's identity on $(q;q)$ in Ramanujan's "most beautiful identity" by Hirschhorn, here is a slightly direct proof of $F_0F_2 + F_1^2=0$.
Recall the Jacobi triple product: $(a^{-1}; q)(aq; q)(q;q)=\sum_n(-1)^na^nq^{n(n+1)/2}$. If $a \neq 1$, then we have $$(a^{-1}q;q)(aq;q)(q;q)=\sum_n(-1)^n\frac{a^n}{1-a^{-1}}q^{n(n+1)/2}.$$ Replacing $a$ by $a^{-1}$, we get $$(aq;q)(a^{-1}q;q)(q;q)=\sum_n(-1)^n\frac{a^{-n}}{1-a}q^{n(n+1)/2}.$$
Averaging the last two identities, we obtain $$(a^{-1}q;q)(aq;q)(q;q)=\sum_n(-1)^n\frac{1}{2}\left(\frac{a^n}{1-a^{-1}}+\frac{a^{-n}}{1-a}\right)q^{n(n+1)/2}.$$
Let $\omega := e^{2\pi i/5}$ be a (primitive) 5th root of unity. Take $a = \omega$, we have
$$(\omega^{-1}q;q)(\omega q;q)(q;q)=\sum_n(-1)^nc_nq^{n(n+1)/2},$$ where $$c_n = \frac{1}{2}\left(\frac{\omega^n}{1-\omega^{-1}}+\frac{\omega^{-n}}{1-\omega}\right) = \begin{cases}\frac{1}{2} & \text{if }n=0\pmod 5 \\ \frac{1+\sqrt{5}}{4} & \text{if }n=1\pmod 5 \\ 0 & \text{if }n=2\pmod 5 \\ -\frac{1+\sqrt{5}}{4} & \text{if }n=3\pmod 5 \\ -\frac{1}{2} & \text{if }n=4\pmod 5\end{cases}.$$
So $(\omega^{-1}q;q)(\omega q;q)(q;q) = \frac{1}{2}(C_0 + \phi_1 C_1 - \phi_1 C_3 - C_4)$, where $C_i = \sum_{n=i\pmod 5}(-1)^nq^{n(n+1)/2}$ and $\phi_1 = \frac{1+\sqrt{5}}{2}$. Notice that by replacing $n$ by $-(n+1)$ in the sums, one can show that $C_0 = -C_4$ and $C_1 = - C_3$. So we get $$(\omega^{-1}q;q)(\omega q;q)(q;q) = C_0+\phi_1 C_1.$$
Similarly, take $a = \omega^2$ and obtain $$(\omega^{-2}q;q)(\omega^2 q;q)(q;q) = \frac{1}{2}(C_0+\phi_2C_1-\phi_2C_3-C_4)=C_0+\phi_2C_1,$$ where $\phi_2 = \frac{1-\sqrt{5}}{2}$. Multiply the last two identities, we have $$(\omega^{-2}q;q)(\omega^{-1}q;q)(\omega q;q)(\omega^{2}q;q)(q;q)(q;q) = C_0^2-C_0C_1-C_1^2.$$
Since $(\omega^{-2}q;q)(\omega^{-1}q;q)(\omega q;q)(\omega^{2}q;q)(q;q) = (q^5;q^5),$ we have $$(q;q)=\frac{C_0^2-C_0C_1-C_1^2}{(q^5;q^5)}.$$ By the same reasoning we used at the very beginning, we have $$F_0 = \frac{C_0^2}{(q^5;q^5)}, F_1 = \frac{-C_0C_1}{(q^5;q^5)}, F_2 = \frac{-C_1^2}{(q^5;q^5)}.$$
Remark: With a bit of extra work (using Jacobi triple product), one can show $C_0 = (q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25}), C_1 = (q^5;q^{25})(q^{20};q^{25})(q^{25};q^{25})$ and hence derive Ramanujan's identity.
