Answer
Eq. (13) cannot be used to evaluate $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $s=1$.
Work Step by Step
We have
$x'\left( s \right) = 1 - \dfrac{1}{{{s^2}}}$, ${\ \ \ }$ $x{\rm{''}}\left( s \right) = \dfrac{2}{{{s^3}}}$,
$y'\left( s \right) = \dfrac{2}{{{s^3}}}$, ${\ \ \ }$ $y{\rm{''}}\left( s \right) = - \dfrac{6}{{{s^4}}}$.
By Eq. (13):
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{x'\left( s \right)y{\rm{''}}\left( s \right) - y'\left( t \right)x{\rm{''}}\left( s \right)}}{{x'{{\left( s \right)}^3}}} = \dfrac{{2 - 6{s^2}}}{{{{\left( { - 1 + {s^2}} \right)}^3}}}$.
Since $x'\left( 1 \right) = 0$, we cannot use Eq. (13) to evaluate $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $s=1$.
