Answer
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = \pi /4}} = - 2\sqrt 2 $
Work Step by Step
We have
$x'\left( t \right) = - \sin \theta $, $x{\rm{''}}\left( t \right) = - \cos \theta $,
$y'\left( t \right) = \cos \theta $, $y{\rm{''}}\left( t \right) = - \sin \theta $.
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}}} = - {\csc ^3}\theta $
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = \pi /4}} = - 2\sqrt 2 $
