1. Given function

$$f(x)= \begin{cases} x^3\sin\left(\frac{1}{x}\right), & x\ne 0 \\ 0, & x=0 \end{cases}$$

We need to check the given statements about $f''(0)$ and $f''\left(\frac{2}{\pi}\right)$.

---

2. First derivative at $x=0$

By definition,

$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h} =\lim_{h\to 0}\frac{h^3\sin(1/h)}{h} =\lim_{h\to 0} h^2\sin(1/h)$$

Since $|\sin(1/h)|\le 1$,

$$|h^2\sin(1/h)|\le h^2 \to 0$$

So,

$$f'(0)=0$$

---

3. Second derivative at $x=0$

Using the definition,

$$f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h} =\lim_{h\to 0}\frac{f'(h)}{h}$$

So first find $f'(x)$ for $x\ne 0$.

For $x\ne 0$,

$$f(x)=x^3\sin\left(\frac{1}{x}\right)$$

Differentiate using product rule:

$$f'(x)=3x^2\sin\left(\frac{1}{x}\right)+x^3\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$

Hence,

$$f'(x)=3x^2\sin\left(\frac{1}{x}\right)-x\cos\left(\frac{1}{x}\right)$$

Therefore,

$$\frac{f'(h)}{h}=3h\sin\left(\frac{1}{h}\right)-\cos\left(\frac{1}{h}\right)$$

Now as $h\to 0$:

• $3h\sin(1/h)\to 0$
• but $\cos(1/h)$ has no limit.

So the limit does not exist.

Hence,

$$f''(0) \text{ does not exist}$$

Therefore:

• A: $f''(0)=0$ is false
• B: $f''(0)=1$ is false

---

4. Second derivative for $x\ne 0$

We already have

$$f'(x)=3x^2\sin\left(\frac{1}{x}\right)-x\cos\left(\frac{1}{x}\right)$$

Differentiate again:

$$f''(x)=\frac{d}{dx}\left(3x^2\sin\left(\frac{1}{x}\right)\right)-\frac{d}{dx}\left(x\cos\left(\frac{1}{x}\right)\right)$$

Now,

$$\frac{d}{dx}\left(3x^2\sin\left(\frac{1}{x}\right)\right) =6x\sin\left(\frac{1}{x}\right)+3x^2\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$ $$=6x\sin\left(\frac{1}{x}\right)-3\cos\left(\frac{1}{x}\right)$$

Also,

$$\frac{d}{dx}\left(x\cos\left(\frac{1}{x}\right)\right) =\cos\left(\frac{1}{x}\right)+x\left[-\sin\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)\right]$$ $$=\cos\left(\frac{1}{x}\right)+\frac{\sin(1/x)}{x}$$

Thus,

$$f''(x)=6x\sin\left(\frac{1}{x}\right)-3\cos\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right)-\frac{\sin(1/x)}{x}$$

So,

$$f''(x)=6x\sin\left(\frac{1}{x}\right)-4\cos\left(\frac{1}{x}\right)-\frac{\sin(1/x)}{x}$$

---

5. Evaluate at $x=\frac{2}{\pi}$

If

$$x=\frac{2}{\pi} \quad \Rightarrow \quad \frac{1}{x}=\frac{\pi}{2}$$

Then,

$$\sin\left(\frac{1}{x}\right)=\sin\left(\frac{\pi}{2}\right)=1, \qquad \cos\left(\frac{1}{x}\right)=\cos\left(\frac{\pi}{2}\right)=0$$

Substitute into $f''(x)$:

$$f''\left(\frac{2}{\pi}\right)=6\left(\frac{2}{\pi}\right)(1)-4(0)-\frac{1}{2/\pi}$$ $$=\frac{12}{\pi}-\frac{\pi}{2}$$

Take LCM $2\pi$:

$$\frac{12}{\pi}-\frac{\pi}{2} =\frac{24-\pi^2}{2\pi}$$

Hence,

$$f''\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2\pi}$$

So:

• C is true
• D is false

---

6. Final conclusion

The only correct option is

$$\boxed{\text{C}}$$