To find the derivative of $y = (8 - 3x^3)^3$ at $x = 2$, we first find the derivative of $y$ with respect to $x$ using the chain rule:

1. Differentiate the outer function: If $u = 8 - 3x^3$, then $y = u^3$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du} = 3u^2$.

2. Differentiate the inner function: The derivative of $u = 8 - 3x^3$ with respect to $x$ is $\frac{du}{dx} = -9x^2$.

3. Apply the chain rule: Combine these derivatives: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot (-9x^2) = -27u^2x^2$.

Now, substitute $x = 2$ into $u$: $u = 8 - 3(2)^3 = 8 - 3 \cdot 8 = 8 - 24 = -16$

Thus, $\frac{dy}{dx}$ at $x = 2$ is: $\frac{dy}{dx} = -27(-16)^2 \cdot (2)^2 = -27 \cdot 256 \cdot 4 = -27648$

Therefore, the derivative of $y$ at $x = 2$ is $-27648$.

Do you need a detailed breakdown or have any other questions related to this? Here are some additional questions you might consider:

1. How would you find the second derivative of the function?
2. How does changing the exponent from 3 to another value affect the derivative?
3. What happens to the derivative as $x$ approaches infinity?
4. How can you confirm this result using a numerical approximation?
5. Would the derivative at $x = 0$ be different, and how would you calculate it?

Tip: When applying the chain rule, always keep track of your inner and outer functions to correctly apply their derivatives.