Answer
Decreasing: $x=-\frac{\sqrt{3}}{3}$
Increasing: $x=\frac{\sqrt{3}}{3}$
Work Step by Step
First derivative
\[
\frac{-2 x}{\left(x^{2}+1\right)^{2}}=y^{\prime}
\]
Second derivative
\[
y^{\prime \prime}=\frac{-3+6 x^{2}}{\left(x^{2}+1\right)^{3}}
\]
\[
\begin{array}{l}
0^{\prime}=\frac{6 x^{2}-2}{\left(1+x^{2}\right)^{3}} \\
-2+6 x^{2}=0 \\
6 x^{2}=2 \\
x=\pm \frac{\sqrt{3}}{3}
\end{array}
\]
The function $y$ increases/decreases most rapidly; if the function $y^{\prime}$ has a maximum/minimum, $y^{\prime \prime}$ has a zero at this value.
\[
y^{\prime}=\frac{-2 \cdot \frac{\sqrt{3}}{3}}{\left(\frac{1}{9}+1\right)^{2}}<0
\]
and
\[
y^{\prime}=\frac{-2 \cdot-\frac{\sqrt{3}}{3}}{\left(1+\frac{1}{9}\right)^{2}}>0
\]
Determine if $y^{\prime}$ is positive or negative at the roots of $y^{\prime \prime}(\text { there, } y$ is increasing most/decreasing most).