To find the next (third) derivative of $y$ with respect to $x$, denoted as $\frac{d^3y}{dx^3}$, we need to take the derivative of $\frac{d^2y}{dx^2}$ with respect to $x$.

Recall that: $\frac{d^2y}{dx^2} = -\frac{1}{32t^3}$

We need to use the chain rule again to differentiate $\frac{d^2y}{dx^2}$ with respect to $x$:

$\frac{d}{dx}\left( \frac{d^2y}{dx^2} \right) = \frac{\frac{d}{dt} \left( \frac{d^2y}{dx^2} \right)}{\frac{dx}{dt}}$

First, compute $\frac{d}{dt} \left( \frac{d^2y}{dx^2} \right)$:

$\frac{d}{dt} \left( -\frac{1}{32t^3} \right) = -\frac{d}{dt} \left( \frac{1}{32t^3} \right) = -\frac{1}{32} \cdot \frac{d}{dt} \left( t^{-3} \right) = -\frac{1}{32} \cdot (-3t^{-4}) = \frac{3}{32t^4}$

Next, divide this by $\frac{dx}{dt} = 2t$:

$\frac{d^3y}{dx^3} = \frac{\frac{d}{dt} \left( \frac{d^2y}{dx^2} \right)}{\frac{dx}{dt}} = \frac{\frac{3}{32t^4}}{2t} = \frac{3}{32t^4} \cdot \frac{1}{2t} = \frac{3}{64t^5}$

Now, evaluate at $t = 2$:

$\frac{d^3y}{dx^3} \Bigg|_{t=2} = \frac{3}{64(2)^5} = \frac{3}{64 \cdot 32} = \frac{3}{2048} = \frac{3}{2048}$

Thus, the third derivative of $y$ with respect to $x$ at $t = 2$ is:

$\frac{d^3y}{dx^3} = \frac{3}{2048}$

Would you like further details or have any questions? Here are some questions you might consider:

1. How do you find higher-order derivatives for parametric equations?
2. Can you explain the process of using the chain rule repeatedly for higher-order derivatives?
3. What are some applications of third derivatives in physics and engineering?
4. How do the first, second, and third derivatives relate to the shape of a graph?
5. Can you provide examples of other parametric equations and their higher-order derivatives?

Tip: When finding higher-order derivatives for parametric equations, carefully apply the chain rule step-by-step and simplify at each stage to avoid errors.