Official solution
x = 3 tan t and y = 3 sec t
So that \[{{dx} \over {dt}}\] = 3sec2t and \[{{dy} \over {dt}}\] = 3 sec t tan t
\[{{dy} \over {dx}}\] = \[{{dy/dt} \over {dx/dt}}\] = sin t
\[{{{d^2}y} \over {d{x^2}}}\] = (cos t) (.{{dt} \over {dx}}\]
\[{{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}\]
\[{{{d^2}y} \over {d{x^2}}}\] = \[{1 \over 3}(cos3 t)
\[{\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}\] = \[{1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}\]