`f(x) = {{:(x^(2)sin'1/x, if x ne 0),(0, if x = 0):}` at `x = 0`.

Explanation

To determine the differentiability of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

at \( x = 0 \), we need to find both the left-hand derivative and the right-hand derivative at that point.

### Step 1: Find the Left-Hand Derivative

The left-hand derivative at \( x = 0 \) is given by:

\[ f'_{-}(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} \]

Substituting \( f(0) = 0 \):

\[ f'_{-}(0) = \lim_{h \to 0^-} \frac{f(h)}{h} = \lim_{h \to 0^-} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} \]

This simplifies to:

\[ f'_{-}(0) = \lim_{h \to 0^-} h \sin\left(\frac{1}{h}\right) \]

Since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, we have:

\[ -h \leq h \sin\left(\frac{1}{h}\right) \leq h \]

As \( h \to 0 \), both bounds approach 0. By the Squeeze Theorem:

\[ f'_{-}(0) = 0 \]

### Step 2: Find the Right-Hand Derivative

The right-hand derivative at \( x = 0 \) is given by:

\[ f'_{+}(0) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} \]

Again substituting \( f(0) = 0 \):

\[ f'_{+}(0) = \lim_{h \to 0^+} \frac{f(h)}{h} = \lim_{h \to 0^+} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} \]

This simplifies to:

\[ f'_{+}(0) = \lim_{h \to 0^+} h \sin\left(\frac{1}{h}\right) \]

Similar to the left-hand case, since \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1:

\[ -h \leq h \sin\left(\frac{1}{h}\right) \leq h \]

As \( h \to 0 \), both bounds approach 0. By the Squeeze Theorem:

\[ f'_{+}(0) = 0 \]

### Step 3: Conclusion

Since both the left-hand and right-hand derivatives at \( x = 0 \) are equal:

\[ f'_{-}(0) = f'_{+}(0) = 0 \]

We conclude that \( f(x) \) is differentiable at \( x = 0 \) and:

\[ f'(0) = 0 \]