Answer
$$\eqalign{
& \left( a \right){\text{Critical points: }}x = 1,\,\,\,x = - 1 \cr
& \left( b \right){\text{Absolute maximum: }}\frac{1}{{16}}{\text{ at }}x = 1 \cr
& \,\,\,\,\,\,\,{\text{Absolute minimum: }} - \frac{1}{{16}}{\text{ at }}x = - 1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{{{\left( {{x^2} + 3} \right)}^2}}}{\text{ on the interval }}\left[ { - 2,2} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{{{{\left( {{x^2} + 3} \right)}^2} - 2x\left( {{x^2} + 3} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( {{x^2} + 3 - 4{x^2}} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( {3 - 3{x^2}} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr
& {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr
& 3 - 3{x^2} = 0 \cr
& 3\left( {1 + x} \right)\left( {1 - x} \right) = 0 \cr
& {\text{Solving this equation for the given interval we obtain}} \cr
& x = 1,\,\,\,x = - 1 \cr
& {\text{Evaluating these point and the endpoints we obtain}} \cr
& f\left( { - 2} \right) = \frac{{ - 2}}{{{{\left( {{{\left( { - 2} \right)}^2} + 3} \right)}^2}}} = - \frac{2}{{49}} \cr
& f\left( { - 1} \right) = \frac{{ - 1}}{{{{\left( {{{\left( { - 1} \right)}^2} + 3} \right)}^2}}} = - \frac{1}{{16}} \cr
& f\left( 1 \right) = \frac{1}{{{{\left( {{{\left( 1 \right)}^2} + 3} \right)}^2}}} = \frac{1}{{16}} \cr
& f\left( 2 \right) = \frac{2}{{{{\left( {{{\left( 2 \right)}^2} + 3} \right)}^2}}} = \frac{2}{{49}} \cr
& {\text{The largest result after evaluating the critical points is }}f\left( 1 \right) = \frac{1}{{16}} \cr
& {\text{which is the absolute maximum}}{\text{.}} \cr
& {\text{The smallest result after evaluating the critical points is }}f\left( { - 1} \right) = - \frac{1}{{16}} \cr
& {\text{which is the absolute minimum}}{\text{.}} \cr} $$
