Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$ 
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.

I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate thrice but it's very lenghty.

Let $n = 15$ and $P(x) = \frac{x^n-1}{x-1} = \sum\limits_{k=0}^{n-1} x^k$. Last part of your question suggest you already know:
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = - \frac12 \left.\frac{d^2}{dx^2} \frac{P'(x)}{P(x)}\right|_{x=1} = -\frac12\left.\frac{d^3}{dx^3}\log P(x)\right|_{x=1}$$
To evaluate the derivative, change variable to $t = \log x$ and let $D$ be the operator $\frac{d}{dt}$, we have
$$-2\mathcal{S} = \left.\left(x^3\frac{d^3}{d x^3}\right)\log P(x)\right|_{x=1}
= D(D-1)(D-2)\left.\log P(e^t)\right|_{t=0}\tag{*1}
$$
Notice
$$\log P(e^t) = \log\frac{e^{nt} - 1}{e^t-1} = \log n + f(nt) - f(t)
\quad\text{ where }\quad f(t) = \log\frac{e^t - 1}{t}$$
We just need to figure out the Taylor expansion of $f(t)$ up to $O(t^4)$. Since
$$f(t) = \log\left( e^{\frac{t}{2}} \frac{\sinh(\frac{t}{2})}{\frac{t}{2}}\right)
= \frac{t}{2} + \log\left( 1 + \frac{t^2}{3! 2^2} + O(t^4)\right)
= \frac{t}{2} + \frac{t^2}{24} + O(t^4)$$
We have
$$Df(t) = \frac12 + \frac{t}{12} + O(t^3)$$
and hence
$$Df(t)|_{t=0} = \frac12,\quad D^2f(t)|_{t=0} = \frac{1}{12}\quad\text{ and }\quad D^3f(t)|_{t=0} = 0$$
Substitute this in $(*1)$, we get
$$-2\mathcal{S} = \bigg[(D^2 - 3D + 2)D(f(nt) - f(t))\bigg]_{t=0} = -\frac{3}{12}(n^2-1) + \frac{2}{2}(n-1)$$
Similplify this give us
$$\sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = \mathcal{S} = \frac{(n-3)(n-1)}{8}$$
For $n = 15$, this reduces to $\displaystyle\;\frac{(15-3)(15-1)}{8} = 21$ as first pointed out by @user64494 in comment.
