$\mathrm{x}=3\mathrm{Tant}, \mathrm{y}=3\mathrm{sect}$ then the value of $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ at $\mathrm{t}=\frac{\mathrm{\pi }}{4}$ is

$$\begin{gathered} \mathrm{x}=3 \mathrm{tant}, \mathrm{y}=3 \mathrm{sect} \\ \frac{\mathrm{dx}}{\mathrm{dt}}=3 {\sec }^2\mathrm{t}, \frac{\mathrm{dy}}{\mathrm{dt}}=3 \mathrm{sect} \mathrm{tant} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}/\mathrm{dt}}{\mathrm{dx}/\mathrm{dt}}=\frac{3 \mathrm{sect} \mathrm{tant}}{3 {\sec }^2 \mathrm{t}}=\frac{\mathrm{tant}}{\mathrm{sect}}=\mathrm{sint} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\mathrm{cost} \frac{\mathrm{dt}}{\mathrm{dx}} \\ =\mathrm{cost} \frac{1}{3 {\sec }^2\mathrm{t}}=\frac{1}{3}{\cos }^3\mathrm{t} \\ {\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}_{\mathrm{t}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{3}{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{1}{6\sqrt{2}} \end{gathered}$$