Given $\left\{\begin{array}{l}x=\frac{1}{3} t^{3} \\ y=\frac{1}{8} t^{8}\end{array}\right.$. Without sliminating the parameter, find $\frac{d^{2} y}{d x^{2}}$ at $t=2$
\[
\left.\frac{d^{2} y}{d x^{2}}\right] t^{2}=
\]

Step 1 ：Given the parametric equations \(x=\frac{1}{3} t^{3}\) and \(y=\frac{1}{8} t^{8}\), we are asked to find \(\frac{d^{2} y}{d x^{2}}\) at \(t=2\) without eliminating the parameter.

Step 2 ：First, we find the first derivative of y with respect to x, which is \(\frac{dy}{dx}\). This can be found by using the chain rule, which states that \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Step 3 ：Next, we find the second derivative of y with respect to x, which is \(\frac{d^2y}{dx^2}\). This can be found by differentiating \(\frac{dy}{dx}\) with respect to t and then dividing by \(\frac{dx}{dt}\).

Step 4 ：Finally, we substitute \(t=2\) into the expression for \(\frac{d^2y}{dx^2}\) to find the value at \(t=2\).

Step 5 ：By substituting \(t=2\) into the expression for \(\frac{d^2y}{dx^2}\), we find that \(\frac{d^2y}{dx^2}\) at \(t=2\) is 20.

Step 6 ：Final Answer: The second derivative of y with respect to x at \(t=2\) is \(\boxed{20}\).