Answer
Substituting $x'\left( t \right)$, $x{\rm{''}}\left( t \right)$, $y'\left( t \right)$ and $y{\rm{''}}\left( t \right)$ into Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = 2$.
Work Step by Step
Let $c\left( t \right) = \left( {t,{t^2}} \right)$. So, we have
$x'\left( t \right) = 1$, $x{\rm{''}}\left( t \right) = 0$,
$y'\left( t \right) = 2t$, $y{\rm{''}}\left( t \right) = 2$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = 2$.
Let $c\left( t \right) = \left( {{t^3},{t^6}} \right)$. So, we have
$x'\left( t \right) = 3{t^2}$, $x{\rm{''}}\left( t \right) = 6t$,
$y'\left( t \right) = 6{t^5}$, $y{\rm{''}}\left( t \right) = 30{t^4}$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = 2$.
Let $c\left( t \right) = \left( {\tan t,{{\tan }^2}t} \right)$. So, we have
$x'\left( t \right) = {\sec ^2}t$, $x{\rm{''}}\left( t \right) = 2{\sec ^2}t\tan t$,
$y'\left( t \right) = 2{\sec ^2}t\tan t$, $y{\rm{''}}\left( t \right) = 2{\sec ^4}t + 4{\sec ^2}t{\tan ^2}t$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = 2$.
