Answer is: option1

\( 4x^2 f(3x^2) + 2g(x^2) \)

Solution:

Let: \[ y = f(x^2) \] Differentiating both sides using the chain rule: \[ \frac{d}{dx} f(x^2) = f'(x^2) \cdot \frac{d}{dx} (x^2) \]

Since \( f'(x) = g(x) \), we substitute \( f'(x^2) = g(x^2) \):

\[ \frac{d}{dx} f(x^2) = g(x^2) \cdot 2x \] Now, differentiate again: \[ \frac{d^2}{dx^2} f(x^2) = \frac{d}{dx} [2x g(x^2)] \] Using the product rule: \[ \frac{d}{dx} [2x g(x^2)] = 2 g(x^2) + 2x \cdot \frac{d}{dx} g(x^2) \] Using the chain rule for \( \frac{d}{dx} g(x^2) \): \[ \frac{d}{dx} g(x^2) = g'(x^2) \cdot \frac{d}{dx} (x^2) = g'(x^2) \cdot 2x \]

Since \( g'(x) = f(3x) \), we substitute \( g'(x^2) = f(3x^2) \):

\[ \frac{d}{dx} g(x^2) = 2x f(3x^2) \]

Thus, the second derivative simplifies to:

\[ \frac{d^2}{dx^2} f(x^2) = 2g(x^2) + 2x(2x f(3x^2)) \] \[ = 2g(x^2) + 4x^2 f(3x^2) \] Final Answer: \[ \boxed{4x^2 f(3x^2) + 2g(x^2)} \]