Second derivative of a logarithmic function via the chain rule

Using the chain rule, $f'(x) = \dfrac{1}{x^2+1}\cdot(2x) = \dfrac{2x}{x^2+1}$. This is a quotient of two functions of $x$, so to differentiate again we need the quotient rule.

With $u = 2x$ and $v = x^2+1$, we have $u' = 2$ and $v' = 2x$. So $f''(x) = \dfrac{u'v - uv'}{v^2} = \dfrac{2(x^2+1) - 2x(2x)}{(x^2+1)^2} = \dfrac{2 - 2x^2}{(x^2+1)^2}$.