Answer
$$\eqalign{
& {\text{Increasing on }}\left( { - 3, - \sqrt 6 } \right),\left( {0,\sqrt 6 } \right) \cr
& {\text{Decreasing on }}\left( { - \sqrt 6 ,0} \right),\,\left( {\sqrt 6 ,3} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2}\sqrt {9 - {x^2}} {\text{ on the interval }}\left( { - 3,3} \right) \cr
& {\text{Derivative}} \cr
& f'\left( x \right) = {x^2}\left( {\frac{{ - 2x}}{{2\sqrt {9 - {x^2}} }}} \right) + 2x\sqrt {9 - {x^2}} \cr
& f'\left( x \right) = \frac{{ - {x^3}}}{{\sqrt {9 - {x^2}} }} + 2x\sqrt {9 - {x^2}} \cr
& {\text{Set the derivative to 0}} \cr
& \frac{{ - {x^3} + 2x\left( {9 - {x^2}} \right)}}{{\sqrt {9 - {x^2}} }} = 0 \cr
& - {x^3} + 18x - 2{x^3} = 0 \cr
& - 3{x^3} + 18x = 0 \cr
& 3x\left( {6 - {x^2}} \right) = 0 \cr
& {\text{Solving the equation we obtain}} \cr
& x = 0,\,\,\,x = - \sqrt 6 {\text{ and }}x = \sqrt 6 \cr
& {\text{From the critical values and the interval }}\left( { - 3,3} \right){\text{ we have}} \cr
& \left( { - 3, - \sqrt 6 } \right),\,\,\left( { - \sqrt 6 ,0} \right),\,\,\left( {0,\sqrt 6 } \right),\,\,\left( {\sqrt 6 ,3} \right) \cr
& {\text{Now}}{\text{, we will evaluate the critical value and resume in a table}} \cr} $$
\[\begin{array}{*{20}{c}}
{{\rm{Interval}}}&{{\rm{Test\, value }}\left( x \right)}&{{\rm{Sign\, of }}f'\left( x \right)}&{{\rm{Behavior\, of }}f\left( x \right)}\\
{\left( { - 3, - \sqrt 6 } \right)}&{ - \frac{5}{2}}& + &{{\rm{Increasing}}}\\
{\left( { - \sqrt 6 ,0} \right)}&{ - 1}& - &{{\rm{Decreasing}}}\\
{\left( {0,\sqrt 6 } \right)}&{\frac{5}{2}}& + &{{\rm{Increasing}}}\\
{\left( {\sqrt 6 ,3} \right)}&1& - &{{\rm{Decreasing}}}\\
{}&{}&{}&{}\\
{}&{}&{}&{}
\end{array}\]
$$\eqalign{
& {\text{From the table we can conlude that the function is:}} \cr
& {\text{Increasing on }}\left( { - 3, - \sqrt 6 } \right),\left( {0,\sqrt 6 } \right) \cr
& {\text{Decreasing on }}\left( { - \sqrt 6 ,0} \right),\,\left( {\sqrt 6 ,3} \right) \cr} $$
