To solve the given problem, we need to find the derivative of $h(x)$ at $x = 1$ given that:

$h(x) = f(2x^3)$

Step-by-Step Solution:

1. Apply the Chain Rule:

   To find $h'(x)$, we use the chain rule:

   $h'(x) = \frac{d}{dx} \left[ f(2x^3) \right] = f'(2x^3) \cdot \frac{d}{dx} (2x^3)$

   Compute the derivative of the inside function:

   $\frac{d}{dx} (2x^3) = 6x^2$

   Thus,

   $h'(x) = f'(2x^3) \cdot 6x^2$

2. Evaluate at $x = 1$:

   Substitute $x = 1$ into the derivative:

   $h'(1) = f'(2 \cdot 1^3) \cdot 6 \cdot 1^2 = f'(2) \cdot 6$

   Given that $f'(2) = 9$, we substitute this value:

   $h'(1) = 9 \cdot 6 = 54$

Final Answer:

$h'(1) = 54$

This is the answer to the problem.

Would you like more details on this problem, or do you have any other questions?

Here are some related questions you might find interesting:

1. How do you apply the chain rule in different contexts?
2. What are the steps to find derivatives of composite functions?
3. How do you interpret the derivative values in real-world applications?
4. What is the importance of differentiability in calculus?
5. How can we extend this problem to multivariable calculus?

Tip: Always remember to apply the chain rule correctly when differentiating composite functions.