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To solve this problem, let's analyze the function \( f(x) = x^2 \sin\left(\sin\left(\frac{1}{x}\right)\right) \) with the condition \( f(0) = 0 \). We need to check which statements (A), (B), or (C) are correct:

### 1. Differentiability Everywhere
To be differentiable everywhere, the function must have a derivative at every point, including at \( x = 0 \).

The behavior of \( f(x) \) around \( x = 0 \) is key. As \( x \) approaches 0, \( \frac{1}{x} \) becomes very large, leading \( \sin(\sin(\frac{1}{x})) \) to oscillate rapidly between -1 and 1. Since \( f(x) \) is multiplied by \( x^2 \), which approaches 0 as \( x \) approaches 0, this might make \( f(x) \) well-behaved at 0.

Let's analyze the derivative:

\[
f'(x) = \frac{d}{dx}\left(x^2 \sin\left(\sin\left(\frac{1}{x}\right)\right)\right).
\]

We need to apply the product rule:
\[
f'(x) = 2x \sin\left(\sin\left(\frac{1}{x}\right)\right) + x^2 \frac{d}{dx}\left(\sin\left(\sin\left(\frac{1}{x}\right)\right)\right).
\]

The derivative \( \frac{d}{dx}\left(\sin\left(\sin\left(\frac{1}{x}\right)\right)\right) \) will involve differentiating \( \sin(\frac{1}{x}) \), which becomes very large and oscillates quickly as \( x \) approaches 0. This results in the derivative not having a limit as \( x \to 0 \), suggesting \( f'(x) \) does not exist at \( x = 0 \). Therefore, \( f(x) \) is **not differentiable everywhere**, eliminating option (A).

### 2. Checking \( f'(0) \neq 0 \) and \( f'(0) = 0 \)
Since we have established that \( f'(x) \) does not exist at \( x = 0 \), this directly implies:

- \( f'(0) \neq 0 \) is **false**.
- \( f'(0) = 0 \) is also **false**.

This eliminates options (B) and (C).

### Conclusion:
Since none of the conditions (A), (B), or (C) are true, the correct answer would be based on verifying the listed options. However, none of the provided combinations match this, which suggests a likely issue with the options given.

Would you like a detailed explanation of the derivatives around \( x = 0 \), or any clarification on specific parts? Here are some related questions:

1. How can we formally prove whether \( f(x) \) is differentiable at \( x = 0 \)?
2. What happens to \( \sin(\sin(\frac{1}{x})) \) as \( x \to 0 \)?
3. Can a function be continuous but not differentiable at a point? Why?
4. How does the behavior of \( x^2 \) near 0 influence the overall function?
5. What techniques are useful in determining the differentiability of a function with oscillatory behavior?

**Tip:** When dealing with oscillatory functions, especially involving terms like \( \frac{1}{x} \), check if the function limits exist and are well-defined, as rapid oscillations can disrupt differentiability.

Which of the following is true for the function f(x) = x^2 sin(sin(1/x)) with f(0) = 0? Options include (A) differentiable everywhere, (B) f'(0) ≠ 0, and (C) f'(0) = 0.

Explore the differentiability of the function f(x) = x^2 sin(sin(1/x)) at x = 0. Analyze whether the function is differentiable everywhere and whether f'(0) exists. This problem involves key calculus concepts such as oscillatory functions and the product rule.

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