A clarification of Ramanujan's theorem. I would like to know if my interpretation is right. Who can give me comfort?
In Notebook IV, Berndt reports Entry 11 (p. 328)
  
and the proof that from, it's purely algebraic. However, I am sure that this theorem is nothing but a modular ninth-degree equation.
The reading key is as follows:
$x=\phi(q)$
$y=\phi(q^{9})$
$z=\phi^{4}(q^{3})$
and the theorem becomes:
"If $ \phi (q) = \phi (q ^ {9}) + \big(\frac{\phi ^ {4} (q ^ {3})} {\phi (q ^ {9})}-\phi ^{3}(q ^ {9})\big) ^ {1/3} $
Then
$3 \phi (q ^ {9}) = \phi (q) + \big (\frac{9 \phi^ {4} (q ^ {3})} {\phi (q)}-\phi^{3} (q ^ {3})\big) ^ {1/3} $ ".
From the third degree modular equation:
$\big(\frac{\phi(q)} {\phi(q^{3})})^{4}=1+2\sqrt{2} \frac{G_{n}^{3}} { G_{9n}^{9}}$
$=\frac{9} {1+2\sqrt{2} \frac{G_{9n}^{3}} { G_{n}^{9}}}$
Substituting the due variables you get the ninth degree modular equation:
$\frac{\phi(q)} {\phi(q^{9})}=1+\sqrt{2} \frac{G_{9n}} { G_{81n}^{3}}$
$3 \frac{\phi(q^{9})} {\phi(q)}=1+\sqrt{2} \frac{G_{9n}} { G_{n}^{3}}$.
$G_{n}, G_{9n}   ,and,  G_{81n}$ are Ramanujan's invariants.
$\phi(q)=1+2\big(q + q^{4} + q^{9} + q^{16}+...+q^{n^{2}})$.
 A: You are right. The first degree 3 modular equation can be written as
$$ \phi(q)^4/\phi(q^3)^4 = 1 + 8\, q\, \chi(q)^3/\chi(q^3)^9.  \tag{1}$$
The dual equation can be written as
$$ 9\, \phi(q^3)^4/\phi(q)^4 = 1 + 8\, \chi(q^3)^3/\chi(q)^9.  \tag{2}$$
The first degree 9 modular equation can be written as
$$ \phi(q)/\phi(q^9) = 1 + 2\, q\, \chi(q^3)/\chi(q^9)^3.        \tag{3}$$
The second degree 9 modular equation can be written as
$$ 3\, \phi(q^9)/\phi(q) = 1 + 2\, \chi(q^3)/\chi(q)^3.           \tag{4}$$
where $\,\chi(q) = (-q,q^2)_{\infty}\,$ is one of Ramanujan's functions.
The two equations in Entry 11 are algebraically equivalent to
$$  \phi(q^3)^4 = \phi(q)^3\phi(q^9) - 3\,\phi(q)^2\phi(q^9)^2 + 3\,\phi(q)\phi(q^9)^3.    \tag{5}$$
and also to Berndt, Notebooks, Part III, p. 218, Equation (24.29).
The resultant of factoring equations $(2)$ and $(4)$, taking the numerators, and eliminating $\,\chi(q)\,$ is the cube of equation $(5)$ muliplied by the cube of $\,72\,(\chi(q^3)\,\phi(q))^3.\,$ This is the algebraic connection.
