Relation between Rogers Ramanujan continued fraction and $j$-invariant While going through this answer I found an interesting but slightly complicated relation between Rogers-Ramanujan continued fraction and the j-invariant. I would like to know an elementary proof of the same.
Before proceeding let me define all the necessary terms and symbolism to set the proper context. Let $\tau$ be a complex number with positive imaginary part and $q=\exp(2\pi i\tau) $ so that $|q|<1$. Below I define functions and the relations which I am aware of. The Rogers-Ramanujan continued fraction is given by $$R(q) =\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\dots}}}}\tag{1}$$ Ramanujan studied this function in great detail and obtained the following fundamental identities $$\frac{1}{R(q)}-1-R(q)=\frac{\eta(q^{1/5})} {\eta(q^5)}\tag{2}$$ and $$\frac{1}{R^5(q)}-11-R^5(q)=\left(\frac{\eta(q)}{\eta(q^5)}\right)^6\tag{3}$$ where $\eta(q) $ is Dedekind eta function defined by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{4}$$ To define the $j$-invariant we need to introduce Ramanujan's version of Eisenstein series denoted by $L, M, N$ (symbols $P, Q, R$ are typically used but we want to avoid conflict with Rogers-Ramanujan continued fraction $R(q) $)
\begin{align}
L(q) &= 1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{5a}\\
M(q)&=1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\tag{5b}\\
N(q) &=1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\tag{5c}
\end{align}
It should be observed that $L$ is related to $\eta$ via $$L(q) =24q\frac{d}{dq}(\log\eta(q))\tag{6}$$ The $j$-invariant is defined as $$j(q) =\frac{1728M^3(q)}{M^3(q)-N^2(q)}\tag{7}$$
Ramanujan obtained a system of differential equations connecting $L, M, N$:
\begin{align}
q\frac{dL(q) } {dq} &=\frac{L^2(q)-M(q)}{12}\tag{8a}\\
q\frac{dM(q)}{dq}&=\frac{L(q)M(q)-N(q)}{3}\tag{8b}\\
q\frac{dN(q)} {dq} &=\frac{L(q) N(q) - M^2(q)}{2}\tag{8c}
\end{align}
Using $(6)$ and $(8a)$ it is evident that $M(q) $ can also be expressed in terms of $\eta(q) $. On the other hand the above differential equations allow us to prove that $$M^3(q)-N^2(q)=1728\eta^{24}(q)\tag{9}$$ and thus we have some expression for $j(q) $ in terms of $\eta(q) $.
The following complicated relation holds between Rogers Ramanujan continued fraction $R(q)$ and $j(q) $ : $$ R^5 (R^{10}+11 R^5-1)^5j+(R^{20}-228 R^{15}+494 R^{10}+228 R^5+1)^3 = 0\tag{10}$$ I checked Wikipedia and found that this is derived from another identity $$j(q) =\frac{(x^2+10x+5)^3} {x} \tag{11}$$ where $$x=125\left(\frac {\eta(q^5)}{\eta(q)}\right)^6\tag{12}$$ Using $(11),(12)$ and $(3)$ we can deduce $(10)$ with a little algebra.
Thus the problem boils down to a proof of equation $(11)$. I don't know if this can be derived using algebraic manipulation of the identities given above or does it need some specific modular equation.
Any proofs or suggestions for proof are welcome. I don't understand the machinery of modular forms properly and would prefer an approach more in the spirit of Ramanujan. The question is however tagged "modular-forms" to get the attention of experts from that tag.

Update: I have finally managed to give a proof based on modular equation of degree $5$ given by Ramanujan and posted it as an answer. The proof is more of a verification and a more natural proof utilizing some transformation formula of eta function is desired.

There is a related question which assumes $(10),(11)$ and proves $(3)$, but the approach uses Mathematica.
 A: (A very extended comment.)
This is not so much answer but to show that equations $(10)$ and $(11)$ are not isolated results, but part of a beautiful family of simple formulas of order $n$. It would be nice to have all these formulas in one place.
Surprisingly, their simplicity depends on basic arithmetic: whether $n$ integrally divides $24/(n-1)$. Note the Dedekind eta function $\eta(q)$ involves the $24$th root $q^{1/24}$. So these are the primes $n = 2,3,5,7,13$ and squares $n=4,9,25$. Let,
$$j(q) =\frac{(x-16)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt2\,\eta(q^2)}{\eta(q)}\right)^{24}$$
$$\; j(q) =\frac{(x+3)^3(x+27)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt3\,\eta(q^3)}{\eta(q)}\right)^{12}$$
$$j(q) =\frac{(x^2+10x+5)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt5\,\eta(q^5)}{\eta(q)}\right)^6$$
$$j(q) =\frac{(x^2+5x+1)^3(x^2+13x+49)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt7\,\eta(q^7)}{\eta(q)}\right)^4$$
$$j(q) =\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x, \;\text{with}\;\; x=\left(\frac {\sqrt{13}\,\eta(q^{13})}{\eta(q)}\right)^2$$
for square orders,
$$j(q) =\frac{(x^2-48)^3}{x^2-64}, \quad\text{with}\quad x=\left(\frac {\sqrt4\,\eta(q^{4})}{\eta(q)}\right)^8+8$$
$$j(q) =\frac{x^3(x^3-24)^3}{x^3-27}, \quad\text{with}\quad x=\left(\frac {\sqrt9\,\eta(q^{9})}{\eta(q)}\right)^3+3$$
$$j(q) = \frac{-(x^{20}+12 x^{15}+14 x^{10}-12 x^5+1)^3}{x^{25} (x^{10}+11 x^5-1)} , \;\text{with}\;\; x^{-1}-x=\left(\frac {\sqrt{25}\,\eta(q^{25})}{\eta(q)}\right)^1+1$$
Alternatively, for square orders $n = 2^2,3^2,5^2$,
$$j(q) =\frac{(x^2+192)^3}{(x^2-64)^2}, \quad\text{with}\quad x=\left(\frac {\eta(q^{1/2})}{\eta(q^2)}\right)^8+8$$
$$j(q) =\frac{x^3(x^3+216)^3}{(x^3-27)^3}, \quad\text{with}\quad x=\left(\frac {\eta(q^{1/3})}{\eta(q^3)}\right)^3+3$$
$$j(q) = \frac{-(r^{20}-228 r^{15}+494 r^{10}+228 r^5+1)^3}{r^5 (r^{10}+11 r^5-1)^5} , \;\text{with}\;\; r^{-1}-r=\left(\frac {\eta(q^{1/5})}{\eta(q^5)}\right)^1+1$$
Order $n=2$ is better known in the guise of the Weber modular functions, while the rest of the first set do not seem to have names.
Generalizing the OP's post, the three square orders can be connected to three kinds of $q$-continued fractions: the first to the octic cfrac (the power $8$ and octahedral symmetry), the second to the cubic cfrac (the power $3$ and tetrahedral symmetry), and the third, of course, to the Rogers-Ramunujan cfrac (and icosahedral symmetry) which are the 3 symmetries of the Platonic solids. Ramanujan studied all three cfracs.
