Let $$f(x) = \begin{cases} x^2\sin\frac{1}{x}; & x \neq 0 \\ 0; & x = 0 \end{cases}$$, then at $$x = 0$$

We need to analyze the continuity and differentiability of $$f(x) = \begin{cases} x^2\sin\frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$$.

We start by examining the continuity of $$f$$ at $$x = 0$$.

$$\lim_{x \to 0} x^2\sin\frac{1}{x} = 0$$ (since $$|x^2\sin(1/x)| \leq x^2 \to 0$$).

This equals $$f(0) = 0$$, so $$f$$ is continuous at $$x = 0$$.

Next, we consider the differentiability of $$f$$ at $$x = 0$$.

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

Therefore, $$f$$ is differentiable at $$x = 0$$ with $$f'(0) = 0$$.

Then, we check the continuity of the derivative $$f'$$ at $$x = 0$$.

For $$x \neq 0$$, $$f'(x) = 2x\sin\frac{1}{x} + x^2 \cdot \cos\frac{1}{x} \cdot \left(-\frac{1}{x^2}\right) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}$$.

We look at the limit:

$$\lim_{x \to 0} f'(x) = \lim_{x \to 0}\left(2x\sin\frac{1}{x} - \cos\frac{1}{x}\right)$$.

The first term tends to 0, but $$\cos(1/x)$$ oscillates between $$-1$$ and $$1$$ as $$x \to 0$$, so this limit does not exist.

Hence, the derivative $$f'$$ is not continuous at $$x = 0$$.

In conclusion, $$f$$ is continuous and differentiable at $$x = 0$$, but $$f'$$ is not continuous there. The correct choice is Option 2: $$f$$ is continuous but $$f'$$ is not continuous.