Answer
$$\eqalign{
& \left( {\text{a}} \right)x = - 3,{\text{ }}x = 0,{\text{ }}x = 3 \cr
& \left( {\text{b}} \right){\text{Decreasing on: }}\left( {0,3} \right),\left( {3,\infty } \right) \cr
& {\text{Increasing on: }}\left( { - \infty , - 3} \right){\text{,}}\left( { - 3,0} \right) \cr
& \left( {\text{c}} \right){\text{relative minimum}}\left( {0,0} \right) \cr} $$
Work Step by Step
$$\eqalign{
& {\text{We have }}f\left( x \right) = \frac{{{x^2}}}{{{x^2} - 9}} \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} - 9}}} \right] \cr
& {\text{By the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} - 9} \right)\left( {2x} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{2{x^3} - 18x - 2{x^3}}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{ - 18x}}{{{{\left( {{x^2} - 9} \right)}^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& - 18x = 0 \cr
& x = 0 \cr
& {\text{The derivative is not defined at }}x = \pm 3,{\text{ so we obtain the critical }} \cr
& {\text{points, }}x = - 3,{\text{ }}x = 0,{\text{ }}x = 3 \cr
& \cr
& \left( {\text{b}} \right) \cr
& {\text{Set the intervals }}\left( { - \infty , - 3} \right),\left( { - 3,0} \right),\left( {0,3} \right),\left( {3,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180}}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty , - 3} \right)}&{\left( { - 3,0} \right)}&{\left( {0,3} \right)}&{\left( {3,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 6}&{x = - 2}&{x = 2}&{x = 6} \\
{{\text{Sign of }}f'\left( x \right)}&{\frac{4}{{27}} > 0}&{\frac{{36}}{{25}} > 0}&{ - \frac{{36}}{{25}} < 0}&{ - \frac{4}{{27}} < 0} \\
{{\text{Conclusion}}}&{{\text{Incr}}{\text{.}}}&{{\text{Incr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}&{{\text{Decr}}{\text{.}}}
\end{array}\]
$$\eqalign{
& f'\left( x \right){\text{ changes from negative to positive at }}x = 0,{\text{ so }}f\left( x \right) \cr
& {\text{has a relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = \frac{{{0^2}}}{{{0^2} - 9}} = 0 \cr} $$
