One nice relation can be written as follow:
$$\sum\limits_{n = - \infty }^ \infty z^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n z^{n^3}= \sum\limits_{n = - \infty }^ \infty (-1)^n z^{2n^3} \phi(z^{6n}) $$
$\quad z \in \mathbb{C},\; |z|=1 $
Where $$\phi(q)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{m^2}$$
More generally if we write
$$\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}= \sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m} z^{2n} q^{2(n^2+m^2)} h^{2n(n^2+3m^2)} \tag{1} $$
The proof of relation (1):
$$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
$$F(-z)=\sum\limits_{n = - \infty }^ \infty (-z)^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-z)^n q^{n^2} h^{n^3}$$
$$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-1)^n z^n q^{n^2} h^{n^3}$$
$$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$
Because of $F(-z)=F(z)$ , we can write that:
$$F(z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)$$
We need to find $A_n(q,h)$
$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
Let's use the transformation
$z=ZQ^{2}h^{3}$
$q=Qh^{3}$
$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n} Q^{2n^2+4n} h^{2n^3+6n^2+6n} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2+2n} h^{n^3+3n^2+3n} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^n Q^{n^2+2n} h^{n^3+3n^2+3n}$$
If we multiply both side by $Z^2Q^2h^2$
$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n+2} Q^{2n^2+4n+2} h^{2n^3+6n^2+6n+2} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1}$$
$$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2(n+1)} Q^{2(n+1)^2} h^{2(n+1)^3} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{(n+1)^2} h^{(n+1)^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{(n+1)^2} h^{(n+1)^3}$$
$$\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{2n} Q^{2n^2} h^{2n^3} A_{n-1}(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n} Q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{n} Q^{n^2} h^{n^3}$$
$$\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{2n} q^{2n^2} h^{2n^3} A_{n-1}(qh^3,h)=\sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{n} q^{n^2} h^{n^3}$$
$$A_n(q,h)=A_{n-1}(qh^3,h)$$
$$A_{-1}(q,h)=A_{0}(qh^{-3},h)$$
$$A_1(q,h)=A_{0}(qh^3,h)$$
$$A_2(q,h)=A_{1}(qh^3,h)=A_{0}(qh^6,h)$$
$$A_{-2}(q,h)=A_{-1}(qh^{-3},h)=A_{0}(qh^{-6},h)$$
$$A_3(q,h)=A_{2}(qh^3,h)=A_{1}(qh^6,h)=A_{0}(qh^9,h)$$
$$A_n(q,h)=A_{0}(qh^{3n},h)$$
We need to find $A_{0}(q,h)$ to complete the proof:
We need to focus on $z^0$ terms
$$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
If we multiply terms by terms for $z^0$ terms
$$A_{0}(q,h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}$$
$$A_{n}(q,h)=A_{0}(qh^{3n},h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}$$
Thus We can write that
$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
$$\sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m}z^{2n} q^{2n^2+2m^2} h^{2n^3+6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
$$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \phi(q^2h^{6n}) =\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$
If we put that $z=1$ and $q=1$
$$\sum\limits_{n = - \infty }^ \infty (-1)^n h^{2n^3} \phi(h^{6n}) =\sum\limits_{n = - \infty }^ \infty h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n h^{n^3} $$
Finally, the same method can be used for $\sum\limits_{n = - \infty }^ \infty z^{n^k}$ to get similiar identies where $k>3$
As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.
Let $p$ be another variable and consider the sum
\begin{align*}
S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm}
\end{align*}
Write it as
\begin{align*}
\sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big)
\end{align*}
and rewrite the inner sums as
\begin{align*}
\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\
&=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\}
\end{align*}
Note that
\begin{align*}
\sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big)
\end{align*}
and that
\begin{align*}
\sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big)
\end{align*}
The remaining sum can be written as
\begin{align*}
\sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\
&=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\
&=-\log\big(1-\frac{w(p)}{w(q)}\big)
\end{align*}
where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}
Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and
\begin{align*}
\exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*}
Now \begin{align*}
\frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q))}\frac{f(w(q))-f(w(p)}{q-p}\\
&=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\
&=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p))}{1-pf^\prime(w(p))} +O(q-p)
\end{align*}
so that $\exp(S(q,q))$ reduces to
$$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q))}\;\;,$$
as desired.
Best Answer
I will try to by more helpful than I was in the comments. First a general observation.
Your infinite product contains socalled Euler functions,
$\phi(q)=\prod_{n=1}^{\infty}(1-q^n)$
which have the Lambert series expansion
$\ln\phi(q)=-\sum_{n=1}^{\infty}\frac{1}{n}\frac{q^{n}}{1-q^{n}}$.
Part of your infinite product can be expanded in this way,
$P(x,y)\equiv\prod_{n=1}^{\infty}\frac{1}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})}= \frac{1}{\phi(\sqrt{xy})\phi(\sqrt{x/y})}=\exp\left[\sum_{n=1}^{\infty}\frac{1}{n}g(x^n,y^n)\right]$.
The function $g(x,y)$ is given by
$g(x,y)=\frac{(xy)^{1/2}}{1-(xy)^{1/2}}+\frac{(x/y)^{1/2}}{1-(x/y)^{1/2}}=\sum_{n=1}^{\infty}x^{n/2}(y^{n/2}+1/y^{n/2}).$
This is not quite what you have written. I have difficulty verifying your expression. For example, the limit $x\rightarrow 0$ of the left-hand side of your expression is $1+\sqrt{x}(\sqrt{y}+1/\sqrt{y})$, but the same limit of the right-hand side is $1+\sqrt{x}(y+1/y)$.
UPDATE 1: Your corrected expression is still problematic; the left-hand side contains the infinite product
$\prod_{n=1}^{\infty}(1+x^n)=\exp\left[-\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}G(x^{n})\right]$, with $G(x)=x/(1-x)$.
The factor $(-1)^n$ in the sum over $n$ seems inconsistent with the right-hand side of your expression
--- but in fact it is consistent (see update 2).UPDATE 2: One more identity is needed, in addition to the Euler function identities, to complete the identification:
$\prod_{n=1}^{\infty}(1+x^n)=\exp\left[\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}\right]$.
We then have, quite generally,
$\prod_{n=1}^{\infty}\frac{(1+x^n)}{(1-q_1^n)(1-q_2^n)}=\exp\left(\sum_{n=1}^{\infty}\frac{1}{n}[F(x^n)+g(q_1^n)+g(q_2^n)]\right)$
with the functions $F(x)=x/(1-x^2)$, $g(q)=q/(1-q)$.
Your equation (1) corresponds to $q_1=\sqrt{x/y}$, $q_2=\sqrt{xy}$.
I know, your function $f(x,y)$ looks much more lengthy, but it is really just $F(x)+g(\sqrt{xy})+g(\sqrt{x/y})$.