Answer
$$\eqalign{
& {\text{Increasing on }}\left( { - 1,0} \right),\left( {1,\infty } \right) \cr
& {\text{Decreasing on }}\left( { - \infty , - 1} \right),\,\left( {0,1} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{2/3}}\left( {{x^2} - 4} \right) \cr
& {\text{Derivative}} \cr
& f'\left( x \right) = {x^{2/3}}\left( {2x} \right) + \frac{2}{3}{x^{ - 1/3}}\left( {{x^2} - 4} \right) \cr
& f'\left( x \right) = 2{x^{5/3}} + \frac{{2\left( {{x^2} - 4} \right)}}{{3{x^{1/3}}}} \cr
& {\text{Set the derivative to 0}} \cr
& 2{x^{5/3}} + \frac{{2\left( {{x^2} - 4} \right)}}{{3{x^{1/3}}}} = 0 \cr
& \frac{{6{x^2} + 2{x^2} - 8}}{{3{x^{1/3}}}} = 0 \cr
& {\text{Is not defined for }}x = 0 \cr
& 8{x^2} - 8 = 0{\text{ }} \cr
& {\text{Solving the equation we obtain}} \cr
& x = - 1,\,\,\,x = 0{\text{ and }}x = 1 \cr
& {\text{From the critical values we can make the next intervals}} \cr
& \left( { - \infty , - 1} \right),\,\,\left( { - 1,0} \right),\,\,\left( {0,1} \right),\,\,\left( {1,\infty } \right) \cr
& {\text{Now}}{\text{, we will evaluate the critical value and resume in a table}} \cr} $$
\[\begin{array}{*{20}{c}}
{{\rm{Interval}}}&{{\rm{Test value }}\left( x \right)}&{{\rm{Sign of }}f'\left( x \right)}&{{\rm{Behavior of }}f\left( x \right)}\\
{\left( { - \infty , - 1} \right)}&{ - 2}& - &{{\rm{Decreasing}}}\\
{\left( { - 1,0} \right)}&{ - \frac{1}{2}}& + &{{\rm{Increasing}}}\\
{\left( {0,1} \right)}&{\frac{1}{2}}& - &{{\rm{Decreasing}}}\\
{\left( {1,\infty } \right)}&2& + &{{\rm{Increasing}}}\\
{}&{}&{}&{}\\
{}&{}&{}&{}
\end{array}\]
$$\eqalign{
& {\text{From the table we can conlude that the function is:}} \cr
& {\text{Increasing on }}\left( { - 1,0} \right),\left( {1,\infty } \right) \cr
& {\text{Decreasing on }}\left( { - \infty , - 1} \right),\,\left( {0,1} \right) \cr} $$
