1. Given function

$$f(x)= \begin{cases} x^2\sin\left(\frac1x\right), & x\ne 0 \\ 0, & x=0 \end{cases}$$

We need to check at $x=0$:

• continuity of $f$
• differentiability of $f$
• continuity of $f'$

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2. Check continuity of $f$ at $x=0$

We compute

$$\lim_{x\to 0} x^2\sin\left(\frac1x\right)$$

Since

$$-1 \le \sin\left(\frac1x\right) \le 1,$$

we get

$$-x^2 \le x^2\sin\left(\frac1x\right) \le x^2.$$

As $x\to 0$,

$$-x^2 \to 0, \qquad x^2 \to 0.$$

So by squeeze theorem,

$$\lim_{x\to 0} x^2\sin\left(\frac1x\right)=0=f(0).$$

Hence, $f$ is continuous at $x=0$.

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3. Check differentiability of $f$ at $x=0$

Using definition,

$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h} =\lim_{h\to 0}\frac{h^2\sin(1/h)}{h} =\lim_{h\to 0} h\sin\left(\frac1h\right).$$

Again,

$$-|h| \le h\sin\left(\frac1h\right) \le |h|,$$

so

$$\lim_{h\to 0} h\sin\left(\frac1h\right)=0.$$

Thus,

$$f'(0)=0.$$

Hence, $f$ is differentiable at $x=0$.

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4. Find $f'(x)$ for $x\ne 0$

For $x\ne 0$,

$$f(x)=x^2\sin\left(\frac1x\right).$$

Differentiate using product rule:

$$f'(x)=2x\sin\left(\frac1x\right)+x^2\cos\left(\frac1x\right)\left(-\frac1{x^2}\right).$$

So,

$$f'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right), \qquad x\ne 0.$$

Also,

$$f'(0)=0.$$

Therefore,

$$f'(x)= \begin{cases} 2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right), & x\ne 0 \\ 0, & x=0 \end{cases}$$

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5. Check continuity of $f'$ at $x=0$

We examine

$$\lim_{x\to 0} f'(x)=\lim_{x\to 0}\left(2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)\right).$$

Now,

$$2x\sin\left(\frac1x\right) \to 0,$$

but

$$\cos\left(\frac1x\right)$$

does not have a limit as $x\to 0$, because it oscillates between $-1$ and $1$.

Hence $f'(x)$ has no limit at $0$.

So $f'$ is not continuous at $x=0$.

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6. Conclusion

• $f$ is continuous at $0$
• $f$ is differentiable at $0$
• $f'$ is not continuous at $0$

Therefore the correct option is:

$$\boxed{\text{A: } f \text{ is continuous but } f' \text{ is not continuous}}$$