Let S be the set of all values of x for which the tangent to the curve

$y=f(x)=x^{3}-x^{2}-2 x$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-2 x-2$

$f(1)=1-1-2=-2, \quad f(-1)=-1-1+2=0$

Since the tangent to the curve is parallel to the line segment joining the points $(1,-2)$ and $(-1,0)$

And their slopes are equal.

$\Rightarrow 3 x^{2}-2 x-2=\frac{-2-0}{2}$

$\Rightarrow x=1, \frac{-1}{3}$

Hence, the required set $S=\left\{\frac{-1}{3}, 1\right\}$