Answer
$y'(1) = 0$
Work Step by Step
$y=\dfrac{1+x+x^2+x^3+x^4+x^5+x^6}{x^3}$
If we divide each term in the numerator by $x^3$, we have:
$y = x^{-3}+x^{-2}+x^{-1}+1+x+x^2+x^3$
Taking the derivative,
$\dfrac{dy}{dx}= (-3)x^{-3-1}+(-2)x^{-2-1}+(-1)x^{-1-1}+0+x^{1-1}+(2)x^{2-1}+(3)x^{3-1} = -3x^{-4}-2x^{-3}-x^{-2}+1+2x+3x^2$
Plugging in $x=1$,
$\dfrac{dy}{dx}$ at $x =1$ is $-3-2-1+1+2+3 = \boxed{0}$