Answer
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = -3}} = 0$
Work Step by Step
We have
$x'\left( t \right) = 8$, $x{\rm{''}}\left( t \right) = 0$,
$y'\left( t \right) = -4$, $y{\rm{''}}\left( t \right) = 0$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = 0$.
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = -3}} = 0$
