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To solve this problem, we are asked to find the second derivative of \( y = e^{x^3} \), which is \( \frac{d^2y}{dx^2} \).

### Step 1: First derivative \( \frac{dy}{dx} \)
We start by finding the first derivative of \( y = e^{x^3} \). Using the chain rule:

\[
\frac{dy}{dx} = \frac{d}{dx} \left( e^{x^3} \right) = e^{x^3} \cdot \frac{d}{dx} (x^3) = e^{x^3} \cdot 3x^2
\]

So, the first derivative is:

\[
\frac{dy}{dx} = 3x^2 e^{x^3}
\]

### Step 2: Second derivative \( \frac{d^2y}{dx^2} \)
Now, we need to differentiate \( \frac{dy}{dx} = 3x^2 e^{x^3} \) again, using the product rule. The product rule states that:

\[
\frac{d}{dx} \left( f(x) g(x) \right) = f'(x) g(x) + f(x) g'(x)
\]

Let \( f(x) = 3x^2 \) and \( g(x) = e^{x^3} \).

- \( f'(x) = \frac{d}{dx} (3x^2) = 6x \)
- \( g'(x) = \frac{d}{dx} \left( e^{x^3} \right) = e^{x^3} \cdot 3x^2 \)

Now applying the product rule:

\[
\frac{d^2y}{dx^2} = 6x e^{x^3} + 3x^2 \cdot (3x^2 e^{x^3})
\]

Simplifying the second term:

\[
3x^2 \cdot 3x^2 = 9x^4
\]

Thus, the second derivative becomes:

\[
\frac{d^2y}{dx^2} = 6x e^{x^3} + 9x^4 e^{x^3}
\]

We can factor out \( e^{x^3} \):

\[
\frac{d^2y}{dx^2} = \left( 6x + 9x^4 \right) e^{x^3}
\]

### Final Answer:
The correct option is:

\[
\boxed{(6x + 9x^4)e^{x^3}}
\]

This corresponds to the first option in the list. 

Would you like further details on this? Here are 5 related questions to deepen your understanding:

1. How do you apply the chain rule to exponential functions?
2. What is the product rule and when do you use it?
3. How can you extend this to higher-order derivatives?
4. What is the general approach for differentiating composite functions like \( e^{x^n} \)?
5. How do you factor out common terms after differentiation?

**Tip**: When applying the product and chain rules, breaking down each term separately can help avoid errors in longer derivatives.

This calculus problem focuses on finding the second derivative of the function y = e^{x^3} using the chain and product rules. Learn step-by-step how to apply these rules to exponential functions. Suitable for Grade 12 or college-level students.

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