A sum involving roots of unity

Here is the proof of Kevin Liu's version $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1 $$ (for the primitive root of unity $y$ of degree $6n+4$) of Nemo's reduction. (Both reductions deserve to be explained, in my opinion.)

We have $$\sum_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}= -\sum_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$

So we should prove

$$2\sum_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$

Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as

$$ 2\sum_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1 $$

Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as $$ \sum_{k=n+1}^{3n+1}\frac{w^{-2k}}{1-w^{-3k}}-\sum_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}= -\sum_{k=n+1}^{3n+1}\frac{w^{k}}{1-w^{3k}}+\sum_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}. $$ Collecting with another half we get (so lucky) $$ \sum_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}= -\sum_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1, $$ since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum_{k=0}^{3n+1} w^{kd}=0$ for all integers $d$ non-divisible by $3n+2$.

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.