Answer
$$\frac{{dy}}{{dt}} = - \frac{6}{{t\sqrt {{t^4} - 9} }}$$
Work Step by Step
$$\eqalign{
& y = {\sin ^{ - 1}}\frac{3}{{{t^2}}} \cr
& y = {\sin ^{ - 1}}\left( {3{t^{ - 2}}} \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {{{\sin }^{ - 1}}\left( {3{t^{ - 2}}} \right)} \right)}}{{dt}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\sin }^{ - 1}}u} \right)}}{{dt}} = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dt}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = 3{t^{ - 2}};{\text{then}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - {{\left( {3{t^{ - 2}}} \right)}^2}} }}\frac{{d\left( {3{t^{ - 2}}} \right)}}{{dt}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{\sqrt {1 - 9{t^{ - 4}}} }}\left( { - 6{t^{ - 2}}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dt}} = - \frac{6}{{{t^2}\sqrt {\frac{{{t^4} - 9}}{{{t^4}}}} }} \cr
& \frac{{dy}}{{dt}} = - \frac{6}{{t\sqrt {{t^4} - 9} }} \cr} $$
