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Let me explainHere is a proof of Nemo's identity. Using the notation $y:=w^{1/2}$, multiplying both sides by $2$, and shifting $k$ by $1$ in three of the five sums, the identity can be rewritten as $$\sum_{k=1}^{2n+1}f_{k,n}(y)=0,$$ where \begin{align*}f_{k,n}(y):=&\frac{(-1)^ky^k}{1-y^{3k}}+\frac{y^k}{1+y^{3k}}-2\frac{(-1)^ky^{k(3k+1)}}{1-y^{6k}}\\[6pt] &+2\frac{(-1)^ky^{(k+1)(3k-2)}}{1-y^{6k-4}}-\frac{(-1)^ky^{(2n+2)(3k-2)}}{1-y^{3k-2}} +\frac{y^{(2n+2)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&(-1)^ky^k\frac{1-y^{3k^2}}{1-y^{3k}}+y^k\frac{1-(-1)^ky^{3k^2}}{1+y^{3k}}\\[6pt] &+(-1)^ky^{(k+1)(3k-2)}\frac{1-y^{(2n-k+1)(3k-2)}}{1-y^{3k-2}} +(-1)^ky^{(k+1)(3k-2)}\frac{1+(-1)^ky^{(2n-k+1)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&\sum_{m=0}^{k-1}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}. \end{align*} In the first $m$-sum, the term $m=k-1$ does not contribute, hence what we really need to prove is $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k=1}^{2n+1}\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)} =0.$$ In the second double sum, we make the change of variables $k':=k+m+1$ and $m':=k-1$. With this notation, the previous equation becomes $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k'=1}^{2n+1}\sum_{m'=0}^{k'-2}\left((-1)^{m'+1}+(-1)^{k'-1}\right)y^{k'(3m'+1)} =0.$$ The two double sums now clearly neutralize each other termwise, and the proof is complete.

Let me explain Nemo's identity. Using the notation $y:=w^{1/2}$, multiplying both sides by $2$, and shifting $k$ by $1$ in three of the five sums, the identity can be rewritten as $$\sum_{k=1}^{2n+1}f_{k,n}(y)=0,$$ where \begin{align*}f_{k,n}(y):=&\frac{(-1)^ky^k}{1-y^{3k}}+\frac{y^k}{1+y^{3k}}-2\frac{(-1)^ky^{k(3k+1)}}{1-y^{6k}}\\[6pt] &+2\frac{(-1)^ky^{(k+1)(3k-2)}}{1-y^{6k-4}}-\frac{(-1)^ky^{(2n+2)(3k-2)}}{1-y^{3k-2}} +\frac{y^{(2n+2)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&(-1)^ky^k\frac{1-y^{3k^2}}{1-y^{3k}}+y^k\frac{1-(-1)^ky^{3k^2}}{1+y^{3k}}\\[6pt] &+(-1)^ky^{(k+1)(3k-2)}\frac{1-y^{(2n-k+1)(3k-2)}}{1-y^{3k-2}} +(-1)^ky^{(k+1)(3k-2)}\frac{1+(-1)^ky^{(2n-k+1)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&\sum_{m=0}^{k-1}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}. \end{align*} In the first $m$-sum, the term $m=k-1$ does not contribute, hence what we really need to prove is $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k=1}^{2n+1}\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)} =0.$$ In the second double sum, we make the change of variables $k':=k+m+1$ and $m':=k-1$. With this notation, the previous equation becomes $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k'=1}^{2n+1}\sum_{m'=0}^{k'-2}\left((-1)^{m'+1}+(-1)^{k'-1}\right)y^{k'(3m'+1)} =0.$$ The two double sums now clearly neutralize each other termwise, and the proof is complete.

Here is a proof of Nemo's identity. Using the notation $y:=w^{1/2}$, multiplying both sides by $2$, and shifting $k$ by $1$ in three of the five sums, the identity can be rewritten as $$\sum_{k=1}^{2n+1}f_{k,n}(y)=0,$$ where \begin{align*}f_{k,n}(y):=&\frac{(-1)^ky^k}{1-y^{3k}}+\frac{y^k}{1+y^{3k}}-2\frac{(-1)^ky^{k(3k+1)}}{1-y^{6k}}\\[6pt] &+2\frac{(-1)^ky^{(k+1)(3k-2)}}{1-y^{6k-4}}-\frac{(-1)^ky^{(2n+2)(3k-2)}}{1-y^{3k-2}} +\frac{y^{(2n+2)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&(-1)^ky^k\frac{1-y^{3k^2}}{1-y^{3k}}+y^k\frac{1-(-1)^ky^{3k^2}}{1+y^{3k}}\\[6pt] &+(-1)^ky^{(k+1)(3k-2)}\frac{1-y^{(2n-k+1)(3k-2)}}{1-y^{3k-2}} +(-1)^ky^{(k+1)(3k-2)}\frac{1+(-1)^ky^{(2n-k+1)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&\sum_{m=0}^{k-1}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}. \end{align*} In the first $m$-sum, the term $m=k-1$ does not contribute, hence what we really need to prove is $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k=1}^{2n+1}\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)} =0.$$ In the second double sum, we make the change of variables $k':=k+m+1$ and $m':=k-1$. With this notation, the previous equation becomes $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k'=1}^{2n+1}\sum_{m'=0}^{k'-2}\left((-1)^{m'+1}+(-1)^{k'-1}\right)y^{k'(3m'+1)} =0.$$ The two double sums now clearly neutralize each other termwise, and the proof is complete.

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GH from MO
GH from MO
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Let me explain Nemo's identity. Using the notation $y:=w^{1/2}$, multiplying both sides by $2$, and shifting $k$ by $1$ in three of the five sums, the identity can be rewritten as $$\sum_{k=1}^{2n+1}f_{k,n}(y)=0,$$ where \begin{align*}f_{k,n}(y):=&\frac{(-1)^ky^k}{1-y^{3k}}+\frac{y^k}{1+y^{3k}}-2\frac{(-1)^ky^{k(3k+1)}}{1-y^{6k}}\\[6pt] &+2\frac{(-1)^ky^{(k+1)(3k-2)}}{1-y^{6k-4}}-\frac{(-1)^ky^{(2n+2)(3k-2)}}{1-y^{3k-2}} +\frac{y^{(2n+2)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&(-1)^ky^k\frac{1-y^{3k^2}}{1-y^{3k}}+y^k\frac{1-(-1)^ky^{3k^2}}{1+y^{3k}}\\[6pt] &+(-1)^ky^{(k+1)(3k-2)}\frac{1-y^{(2n-k+1)(3k-2)}}{1-y^{3k-2}} +(-1)^ky^{(k+1)(3k-2)}\frac{1+(-1)^ky^{(2n-k+1)(3k-2)}}{1+y^{3k-2}}\\[6pt] =&\sum_{m=0}^{k-1}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)}. \end{align*} In the first $m$-sum, the term $m=k-1$ does not contribute, hence what we really need to prove is $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k=1}^{2n+1}\sum_{m=0}^{2n-k}\left((-1)^k+(-1)^{k+m}\right)y^{(k+m+1)(3k-2)} =0.$$ In the second double sum, we make the change of variables $k':=k+m+1$ and $m':=k-1$. With this notation, the previous equation becomes $$\sum_{k=1}^{2n+1}\sum_{m=0}^{k-2}\left((-1)^k+(-1)^m\right)y^{k(3m+1)}+ \sum_{k'=1}^{2n+1}\sum_{m'=0}^{k'-2}\left((-1)^{m'+1}+(-1)^{k'-1}\right)y^{k'(3m'+1)} =0.$$ The two double sums now clearly neutralize each other termwise, and the proof is complete.