Answer
$$\eqalign{
& {\text{Increasing on }}\left( { - \sqrt 2 ,\sqrt 2 } \right) \cr
& {\text{Decreasing on }}\left( { - \infty , - \sqrt 2 } \right),\,\left( {\sqrt 2 ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{x}{{{x^2} + 2}}} \right) \cr
& {\text{Derivative}} \cr
& f'\left( x \right) = \frac{1}{{1 + {{\left( {x/\left( {{x^2} + 2} \right)} \right)}^2}}}\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 2}}} \right] \cr
& f'\left( x \right) = \frac{{{{\left( {{x^2} + 2} \right)}^2}}}{{{{\left( {{x^2} + 2} \right)}^2} + {x^2}}}\left( {\frac{{{x^2} + 2 - x\left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^2}}}} \right) \cr
& f'\left( x \right) = \frac{{2 - {x^2}}}{{{{\left( {{x^2} + 2} \right)}^2} + {x^2}}} \cr
& {\text{Set the derivative to 0}} \cr
& 2 - {x^2} = 0 \cr
& {\text{Solving the equation we obtain}} \cr
& x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr
& {\text{From the critical values and the domain }}\left( { - \infty ,\infty } \right){\text{ we have}} \cr
& \left( { - \infty , - \sqrt 2 } \right),\,\,\left( { - \sqrt 2 ,\sqrt 2 } \right),\,\,\left( {\sqrt 2 ,\infty } \right) \cr
& {\text{Now}}{\text{, we will evaluate the critical value and resume in a table}} \cr} $$
\[\begin{array}{*{20}{c}}
{{\rm{Interval}}}&{{\rm{Test\ value }}\left( x \right)}&{{\rm{Sign\ of }}f'\left( x \right)}&{{\rm{Behavior\ of }}f\left( x \right)}\\
{\left( { - \infty , - \sqrt 2 } \right)}&{ - 2}& - &{{\rm{Decreasing}}}\\
{\left( { - \sqrt 2 ,\sqrt 2 } \right)}&0& + &{{\rm{Increasing}}}\\
{\left( {\sqrt 2 ,\infty } \right)}&2& - &{{\rm{Decreasing}}}\\
{}&{}&{}&{}
\end{array}\]
$$\eqalign{
& {\text{From the table we can conlude that the function is:}} \cr
& {\text{Increasing on }}\left( { - \sqrt 2 ,\sqrt 2 } \right) \cr
& {\text{Decreasing on }}\left( { - \infty , - \sqrt 2 } \right),\,\left( {\sqrt 2 ,\infty } \right) \cr} $$
