Sum of complex roots' fractions

Note that $\omega, \ldots, \omega^6$ are precisely the roots of the sixth degree polynomial: $$p(x) = x^6 + \cdots + 1 = \dfrac{x^7-1}{x-1}.$$ Thus, we can write $$p(x) = (x-\omega)\cdots(x-\omega^6).$$ Taking (natural) $\log$ on both sides and differentiating gives us $$\dfrac{p'(x)}{p(x)} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$ Note that \begin{align} \log p(x) &= \log(x^7 - 1) - \log(x-1)\\ \implies \dfrac{p'(x)}{p(x)} &= \dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1}. \end{align}

This gives us that $$\dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$

Differentating both sides again gives us $$\dfrac{(x^7-1)(42x^5) - (7x^6)(7x^6)}{(x^7-1)^2} + \left(\dfrac{1}{x-1}\right)^2 = -\left(\dfrac{1}{x-\omega}\right)^2-\cdots-\left(\dfrac{1}{x-\omega^6}\right)^2.$$

Now, we simply substitute $x = -1$ both sides. It is clear that the RHS transforms to the negative of what we want, whereas the LHS becomes \begin{align} \dfrac{(-2)(-42) - (7)(7)}{(-2)^2} + \left(\dfrac{1}{-2}\right)^2 &= \dfrac{84-49}{4} + \dfrac{1}{4}\\ &= \dfrac{36}{4} = 9 \end{align}

This gives us the answer as $-9$.