Answer
$-\frac{6}{{t}\sqrt( { {t^4}-9})}$
Work Step by Step
Given that $y$ = $sin^{-1}{\frac{3}{t^2}}$
on substituting ${\frac{3}{t^2}}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dsin^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{u^2})}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $\frac{1}{(\sqrt ({1-{(\frac{3}{t^2})^2}})}$ $\times$ $\frac{d {(\frac{3}{t^2})}}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt( {1-(\frac{9}{t^4})})}$ $\times$ $-\frac{6}{t^3}$
The final answer is: $-\frac{6}{{t}\sqrt( { {t^4}-9})}$