As with Mariano, I too have no idea what sort of mathematical sorcery Ramanujan used, but what you have there is a special case of the Rogers-Ramanujan continued fraction (sans the $\sqrt[5]{q}$ factor).
Letting
$$R(q)=\cfrac{\sqrt[5]{q}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cdots}}}$$
the question amounts to how one might arrive at
$$R(\exp(-2\pi))=\sqrt{\phi\sqrt{5}}-\phi$$
As noted in this paper, Ramanujan had derived an explicit formula for $R(q)$ for certain special values of $q$ in his "lost notebook":
$$\begin{align*}R\left(\exp(-2\pi\sqrt{n})\right)&=\frac1{4t_n^2}\left((1-\phi t_n)\sqrt{1-t_n}-\sqrt{(1-t_n)(1+\phi t_n)^2-4\phi t_n}\right)\times \\&\left(\sqrt{(1-t_n)\left(1-\frac{t_n}{\phi}\right)^2+\frac{4 t_n}{\phi}}-\left(1+\frac{t_n}{\phi}\right)\sqrt{1-t_n}\right)\end{align*}$$
where $t_n=-\dfrac{G_{n/25}}{G_{25n}}$ and $G_n$ is the Ramanujan $G$-function (a.k.a. Ramanujan(-Weber) class invariant),
$$G_n=2^{-\frac14}q^{-\frac1{24}}\prod_{k=0}^\infty (1+q^{2k+1})=2^{-\frac14}q^{-\frac1{24}}(-q;q^2)_\infty$$
and $q=\exp(-\pi\sqrt n)$. (the details for proving it are a bit involved, but they can be seen in the linked paper, or here.)
Your case corresponds to $n=1$. Since $G_{1/n}=G_{n}$ (proven in this paper), $t_n=-1$, and making this substitution into the special formula for $R\left(\exp(-2\pi\sqrt{n})\right)$ yields the desired identity.
