What will you try?

Let $\omega, \omega^2$ denote the complex cube roots of unity such that

$\omega = \frac{-1+i\sqrt{3}}{2}$

$\omega^2 = \frac{-1-i\sqrt{3}}{2}$

Therefore

$\large f(x) = \frac{1}{(x+\omega)(x+\omega^2)} = \frac{1}{\omega^2-\omega}\left(\frac{1}{x+\omega}-\frac{1}{x+\omega^2}\right)$ .

We know that the $n^{th}$ derivative of $\frac{1}{x}$ is $\frac{(-1)^nn!}{x^{n+1}}$

Therefore, we can write

$\large f_6(x) = \frac{6!}{\omega^2-\omega}\left(\frac{1}{(x+\omega)^7} - \frac{1}{(x+\omega^2)^7}\right) = \frac{6!}{\omega^2-\omega }\left(\frac{(x+\omega^2)^7-(x+\omega)^7}{(x+\omega)^7(x+\omega^2)^7}\right)$

$x+\omega^2 = \left(x-\frac{1}{2}\right) + i\left(-\frac{\sqrt{3}}{2}\right) = r(cos\theta - isin\theta)$ ... $(1)$

$x+\omega = \left(x-\frac{1}{2}\right) + i\left(\frac{\sqrt{3}}{2}\right) = r(cos\theta + isin\theta)$ .... $(2)$

Therefore,

$(x+\omega^2)^7 = r^7(cos7\theta-isin7\theta)$ ...... $(3)$

$(x+\omega)^7 = r^7(cos7\theta+isin7\theta)$ ......... $(4)$

$\omega^2 - \omega = -i\sqrt{3}$ ............................... $(5)$

Therefore from equations $(3)$ , $(4)$ and $(5)$ we can write,

$\large f_6(x) = \frac{6!}{-i\sqrt{3}}\left(\frac{r^7(-2isin7\theta)}{r^{14}}\right) = \frac{6!}{\sqrt{3}}\frac{2sin7\theta}{r^7}$

Now multiplying and dividing by $sin^7\theta$ we get,

$\large f_6(x) = \frac{(6!)(2)}{\sqrt{3}}\frac{sin7\theta\times sin^7\theta}{(rsin\theta)^7}$ ...... $(6)$

From equations $(1)$ and $(2)$ we can write

$(x+\omega)-(x+\omega^2) = 2ri sin\theta$

$\omega-\omega^2 = 2ri sin\theta$

$\left(\frac{\sqrt{3}}{2}\right)^7 = (rsin\theta)^7$

Now substituting value of $(rsin\theta)^7$ from above in equation $(6)$ we get,

$\large f_6(x) = \frac{(6!)(2)^8}{(\sqrt{3})^8}(sin^7\theta)(sin7\theta)$

But from equation $(1)$ (or even $(2)$ ) we can write $\theta = arctan\left(\frac{\sqrt{3}}{2x-1}\right)$

hence,

$\boxed{\large f_6(x) = \frac{20480}{9} \sin^7 \left( \text{arctan} \left( \frac {\sqrt 3}{2x-1} \right) \right) \sin \left( 7 \cdot \text{arctan} \left( \frac {\sqrt 3}{2x-1} \right) \right)}$