If y=1/(x+1), then (d^2y)/(dx^2) at x = 2 is:

$y=\frac{1}{x+1}$ so differentiating y w.r.t. x

$⇒\frac{dy}{dx}=\frac{-1}{(x+1)^2}$ again differentiating w.r.t x

$⇒\frac{d^2y}{dx^2}=\frac{2}{(x+1)^3}$ so $\frac{d^2y}{dx^2}|_{x=2}=\frac{2}{(x+1)^3}=\frac{2}{3^3}$

$=\frac{2}{27}$