Answer
$f(x)$ increasing on $\left( -\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right )$ and decreasing on $\left(-\infty, -\frac{1+\sqrt{5}}{2}\right )\cup\left( \frac{1+\sqrt{5}}{2},\infty\right ) $
$f(x)$ has a local maximum at $x= \frac{1+\sqrt{5}}{2} $ and a local minimum at $x=- \frac{1+\sqrt{5}}{2} $.
Work Step by Step
Given $$y=\frac{2 x+1}{x^{2}+1}$$
Since
\begin{align*}
f'(x) &=\frac{\frac{d}{dx}\left(2x+1\right)\left(x^2+1\right)-\frac{d}{dx}\left(x^2+1\right)\left(2x+1\right)}{\left(x^2+1\right)^2}\\
&= \frac{-2x^2-2x+2}{\left(x^2+1\right)^2}
\end{align*}
Then $f'(x)=0$ for $-2x^2-2x+2= 0 $. Hence $$\quad x=-\frac{1+\sqrt{5}}{2},\:x=\frac{\sqrt{5}-1}{2} $$
Choose $x=-2,\ \ x=0 $ and $x=1$:
\begin{align*}
f'(-2)&< 0\\
f'(0)&>0\\
f'(1)&<0
\end{align*}
Then $f(x)$ is increasing on $\left( -\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right )$ and decreasing on $\left(-\infty, -\frac{1+\sqrt{5}}{2}\right )\cup\left( \frac{1+\sqrt{5}}{2},\infty\right ) $. Hence, $f(x)$ has a local maximum at $x= \frac{1+\sqrt{5}}{2} $ and a local minimum at $x=- \frac{1+\sqrt{5}}{2}$.
