Answer
$$\frac{d^2y}{dx^2}=\frac{4}{(1-x)^3}$$
Work Step by Step
$$\frac{dy}{dx}=\big(\frac{1+x}{1-x}\Big)'=
\frac{(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^2}=
\frac{1-x-(1+x)\cdot(-1)}{(1-x)^2}=\frac{2}{(1-x)^2}$$
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=\Big(\frac{2}{(1-x)^2}\Big)'=(2(1-x)^{-2})'=2\cdot(-2)(1-x)^{-3}(1-x)'=-\frac{4}{(1-x)^3}\cdot(-1)=\frac{4}{(1-x)^3}$$
