Let $e(n)$ be the number of partitions of $n$ with even number of even parts and let $o(n)$ denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that $\underset{n\u0101\x89\u201e0}{\u0101\x88\x91}(e(n)\u0101\x88\x92o(n)){x}^{n}=\frac{1}{(1\u0101\x88\x92x)\u0106\x97(1+{x}^{2})\u0106\x97(1\u0101\x88\x92{x}^{3})\u0106\x97(1+{x}^{4})\u0106\x97...}$. I have been racking my head over this for the past few hours, and I can't see any light.

Noticed, that $e(n)\u0101\x88\x92o(n)=2e(n)\u0101\x88\x92p(n)$ where $p(n)$ is the number of partitions of $n$, so the above claim is equivalent to showing $\underset{n\u0101\x89\u201e0}{\u0101\x88\x91}e(n){x}^{n}=\frac{1}{2}\frac{1}{(1\u0101\x88\x92x)(1\u0101\x88\x92{x}^{3})(1\u0101\x88\x92{x}^{5})...}(\frac{1}{(1\u0101\x88\x92{x}^{2})(1\u0101\x88\x92{x}^{4})....}+\frac{1}{(1+{x}^{2})(1+{x}^{4})........})$, and similarly, it is equivalent to $\underset{n\u0101\x89\u201e0}{\u0101\x88\x91}o(n){x}^{n}=\frac{1}{2}\frac{1}{(1\u0101\x88\x92x)(1\u0101\x88\x92{x}^{3})(1\u0101\x88\x92{x}^{5})...}(\frac{1}{(1\u0101\x88\x92{x}^{2})(1\u0101\x88\x92{x}^{4})....}\u0101\x88\x92\frac{1}{(1+{x}^{2})(1+{x}^{4})........})$, but these identities appear more difficult than the original one.