By the product rule,
[tex]\dfrac{dy}{dx} = \dfrac{x-1}{x+1} \dfrac{d(7x^7-x^2)}{dx} + \dfrac{7x^7-x^2}{x+1} \dfrac{d(x-1)}{x+1} + (7x^7-x^2)(x-1) \dfrac{d\left(\frac1{x+1}\right)}{dx}[/tex]
By the power and chain rules,
[tex]\dfrac{dy}{dx} = \dfrac{(x-1)(49x^6-2x)}{x+1} + \dfrac{7x^7-x^2}{x+1} - \dfrac{(7x^7-x^2)(x-1)}{(x+1)^2}[/tex]
When [tex]x=1[/tex], the first and last term vanish, and we're left with
[tex]\dfrac{dy}{dx}\bigg|_{x=1} = \dfrac{7 - 1}{1 + 1} = \dfrac62 = \boxed{3}[/tex]
