In his paper On certain Arithmetical Functions (published in Transactions of the Cambridge Philosophical Society, XXII, No. 9, 1916, pp. 159-184) Ramanujan presents the following identities (as if they are quite well known and rather obvious)
\begin{align}
f(q) &= q^{1^2/24}-q^{5^2/24}-q^{7^2/24}+q^{11^2/24}+\dots\tag{1}\\
f^3(q)&=q^{1^2/8}-3q^{3^2/8}+5q^{5^2/8}-7q^{7^2/8}+\dots\tag{2}\\
\frac{f^5(q)}{f^2(q^2)}&=q^{1^2/24}-5q^{5^2/24}+7q^{7^2/24}-11q^{11^2/24}+\dots\tag{3}\\
\frac{f^5(q^2)}{f^2(-q)}&=q^{1^2/3}-2q^{2^2/3}+4q^{4^2/3}-5q^{5^2/3}+\dots\tag{4}
\end{align}
where $$f(q) =q^{1/24}(1-q)(1-q^2)(1-q^3)\dots\tag{5}$$ and $1,2,4,5, \dots$ are the natural numbers without the multiples of $3$ and $1,5,7,11,\dots$ are the natural odd numbers without the multiples of $3$.
The function $f(q) $ is Dedekind's eta function and due to the presence of 24th root the symbol $f(-q) $ is confusing. What is really meant by $f(-q) $ is the expression $$q^{1/24}(1-(-q))(1-(-q)^2)(1-(-q)^3)\dots$$ so that we change the sign of $q$ in factors other than $q^{1/24}$.
The formulas $(1),(2)$ are really well known as Euler's pentagonal theorem and Jacobi identity respectively and both of these can be derived using the Jacobi triple product identity $$\prod_{n=1}^{\infty}(1-q^{2n})(1+q^{2n-1}z)(1+q^{2n-1}z^{-1})=\sum_{n=-\infty}^{\infty}z^nq^{n^2}\tag{6}$$
Formula $(3)$ was recently discussed in this question and a proof based on Quintuple product identity is given in one of the answers.
The formula $(4)$ is the one which is left to prove and the similarity of its left hand side with that of $(3)$ indicates that it may be amenable to Quintuple product identity as well. However I have been unable to establish it so far using Quintuple product identity.
The identity is given in the form $$\frac{(\eta(3z)\eta(12z))^2}{\eta(6z)}=\sum_{n=1}^{\infty}\left(\frac{n}{3}\right) nq^{n^2}\tag{7}$$ (where $q=\exp(2\pi i z) $ and $\eta(z) =f(q)$) as a part of a larger set of identities in the paper Eta-quotients and theta functions by Robert J. Lemke Oliver and it is established using the theory of modular forms.
I am searching for a proof which avoids modular forms and is of a more elementary character.
Best Answer
As you suspected, (4) can be proved from the quintuple product identity $$ \prod_{n\ge 1} (1-s^n)(1-s^nt)(1-s^{n-1}t^{-1})(1-s^{2n-1}t^2)(1-s^{2n-1}t^{-2}) = \sum_{n}s^{(3n^2+n)/2}(t^{3n}-t^{-3n-1}). $$ This can be done as follows: set $s:=q^2$ and $t:=q^{-1}y$; then you get \begin{eqnarray*} &\ &\prod_{n\ge 1} (1-q^{2n})(1-q^{2n-1}y)(1-q^{2n-1}y^{-1})(1-q^{4n-4}y^2)(1-q^{4n}y^{-2})\\ &=&\sum_{n}q^{3n^2-2n} y^{3n} - q^{3n^2+4n+1} y^{-3n-1} \end{eqnarray*} so, shifting the index $n$ in two places, \begin{eqnarray*} &(1-y^2)&\prod_{n\ge 1} (1-q^{2n})(1-q^{2n-1}y)(1-q^{2n-1}y^{-1})(1-q^{4n}y^2)(1-q^{4n}y^{-2})\\ &=&\sum_{n}q^{3n^2-2n} (y^{3n} - y^{-(3n-2)}). \end{eqnarray*} Divide both sides by $1-y^2$ and then set $y:=1$; this gives \begin{eqnarray*} \prod_{n\ge 1} (1-q^{2n})(1-q^{2n-1})^2(1-q^{4n})^2 &=&-\sum_{n} (3n-1)q^{3n^2-2n}\\ &=&\sum_n (3n+1)q^{3n^2+2n}\\ &=&\sum_{n\ge 1} (\frac n 3) n q^{(n^2-1)/3}, \end{eqnarray*} which becomes (4) after multiplying by $q^{1/3}$ and rewriting the left-hand side. This proof is more or less the same as the one given by Basil Gordon in 1961 [1].
1: (12) in "Some Identities in Combinatorial Analysis", B. Gordon, The Quarterly Journal of Mathematics, 12, #1 (1961), pp. 285-290.