Are the values I calculated correct?
$S(q)=-R(-q)$ where R(q) is the Rogers-Ramanujan continued fraction. $R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$
The relationship from which the solution comes is: $$\frac{1}{S^{5}(q)}+11-S^{5}(q)=D(n)$$
Setting $D(n)=\frac{f^{6}(q)}{q.f^{6}(q^{5})}=11-2c$ with $q=e^{-\pi\sqrt{n}}$
$$S(e^{-\pi\sqrt{n}})=(c+\sqrt{c^{2}+1})^{1/5}$$ Therefore $D(n)=\frac{f^{6}(q)}{q.f^{6}(q^{5})}=(\frac{G_{25n}}{G_{n}})^{6}(1+2\frac{G_{n}}{G_{25n}^{5}})^{3}$
The values $G_{25}=\varphi$ and $G_{625}$ are taken from Ramanujan's Class Invariant $G_{625}$
$D(25)=\frac{125}{64}\frac{\varphi^{15}}{A^{6}}(-\varphi^{3}+A^{2})^{6}$
$D(\frac{1}{625})=\frac {64 A^{6}} {\varphi^{15}(-\varphi^{3}+A^{2})^{6}}$
where
$\varphi$ is the golden ratio
$A=1+(4\varphi)^{1/5}\big((3+\frac{5^{1/4}}{\varphi^{3/2}})^{1/5}+ (3-\frac{5^{1/4}}{\varphi^{3/2}})^{1/5}\big)$
$$S(e^{-5\pi})=\frac{1}{2}-\frac{\sqrt{5}}{2}+\frac{\sqrt{5}}{1+\big(\frac{37}{2}+\frac{17\sqrt{5}}{2}+\sqrt{765+342\sqrt{5}}\big)^{1/5}}$$
Approximate values are
$S(e^{-5\pi})≈0.043213924776185367286077818593870276674898615136$
and
$S(e^{-\frac{\pi}{25}})≈1.618033443506055401293042879331980849826395981166$
A similar example is:
$n=5:$
$$D(5)=\frac{5\sqrt{5}\varphi^{6}(-\sqrt{5}+ A^{2})^{6}}{64 A^{6}}$$,
$A=1+\frac{4^{1/5}}{\sqrt{\varphi}}\big(a^{1/5}+b^{1/5})$
$a=(4-\sqrt{5})\varphi^{3/2}+5^{3/4}$
$b=(4-\sqrt{5})\varphi^{3/2}-5^{3/4}$
$$S(e^{-\pi\sqrt{5}})= 0.2455946125706676477155956393239719960185372$$.
$$D(\frac{1}{125})=\frac{64.5.\sqrt{5}A^{6}} {\varphi^{6}(-\sqrt{5}+A^{2})^{6}}$$
$$S(e^{-\frac{\pi}{5\sqrt{5}}})= 1.614820256133332390035507688296554012559141$$
