Fix some sequence $(a_k)$ such that $a_k=R_k+q^k$ and $R_k=\frac{1}{R(q^k,q)}$,where $R(x,q)$ is a Generalized Rogers-Ramanujan continued fraction for complex number $x$ and $q=\exp{(2\pi i\tau)}$ is the nome,$|q|\lt1$,and consider $$F(q)=\left(\sum_{n=1}^{\infty} \frac{(-1)^n n}{A_{n-1}}q^{n(n+1)/2}\right)^2- \sum_{n=1}^{\infty} \left(\frac{(-1)^n n}{A_{n-1}}q^{n(n+1)/2}\right)^2\tag1$$ which is satisfied by identity(3) in the other post $$\frac{(q)^5_{\infty}}{(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty} \frac{(-1)^n n}{A_{n-1}} q^{n(n+1)/2}$$ where $$A_n=\prod_{k=1}^{n-1} a_{k}$$ On the other hand, consider, for every positive $k$,$$ G_k(q)=\frac{k}{A_{k-1}}\sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{A_{n+k-2}}(n+k-1)q^{(n^2+(2k-1)n+2k^2)/2}\tag2$$
How to show that $$F(q)=2\sum_{k=1}^\infty G_k(q)\ ?$$
As other commenters suggested,I'll provide a motivating special case(around the boundary of convergence of the Generalized Rogers-Ramanujan continued fraction)
If we consider the fact that $\lim_{x\rightarrow 0+}\frac{1}{R(e^{-nx},e^{-x})}=\phi$ where $\phi=\frac{1+\sqrt5}{2}$ is the golden ratio and applying our conjectured identity,we are led to $$F(1)=\Big(\sum_{n=1}^{\infty}\frac{(-1)^n n}{(1+\phi)^n}\Big)^2-\sum_{n=1}^{\infty}\Big(\frac{(-1)^n n}{(1+\phi)^n}\Big)^2=\frac{1}{25}(1-3\sqrt5)$$
and
$$F(1)=\sum_{k=1}^{\infty}\Big(\sum_{n=2}^{\infty} \frac{(-1)^{n-1}}{(1+\phi)^{n+2k-1}}(2k)(n+k-1)\Big)=\frac{1}{25}(1-3\sqrt5)$$
