Answer
$f''(1) = \frac{1}{4}$
Work Step by Step
$f'(x) = \frac{2x(1+x) - (x^2)(1)}{(1+x)^2}$
$f'(x) = \frac{2x+x^2 - x^2}{(1+x)^2}$
$f'(x) = \frac{2x+x^2}{(1+x)^2}$
$f''(x) = \frac{(1+x)^2(2+2x)-[(-2x+x^2)(2(1+x))}{((1+x)^2)^2}$
$f''(x) = \frac{(1+x)^2(2+2x)-2(2x+x^2)}{(1+x)^4}$
Cancel out one $(1+x)$:
$f''(x) = \frac{(1+x)(2+2x)-2(2x+x^2)}{(1+x)^3}$
$f''(1) = \frac{(1+1)(2+2(1))-2(2(1)+(1)^2)}{(1+1)^3}$
$f''(1) = \frac{2(4)-2(2+1)}{(2)^3}$
$f''(1) = \frac{8-6}{8}$
$f''(1) = \frac{2}{8}$
$f''(1) = \frac{1}{4}$