Answer
Yes , the function $g$ is differentiable at $x = 0, g'(0) = 0.$
Work Step by Step
$f$ is continuous at $x = 0$. Using the alternative form of the derivate, you have:
$$\lim_{x \to 0} \frac{f(x) - f(c)}{x - c}$$
$$\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$$
$$\lim_{x \to 0} \frac{x \sin(1/x) - 0}{x - 0}$$
$$\lim_{x \to 0} {\left(\sin \frac 1x\right)}$$
Because this limit does not exist $( \sin(1/x)$ oscillates between $–1$ and $1$), the function is not differentiable at $x = 0$
and $g$ is continuous at $x = 0$. Using the alternative form of the derivative again, you have:
$$\lim_{x \to 0} \frac{g(x) - g(c)}{x - c}$$
$$\lim_{x \to 0} \frac{g(x) - g(0)}{x - 0}$$
$$\lim_{x \to 0} \frac{x^{2} \sin(1/x) - 0}{x - 0}$$
$$\lim_{x \to 0} {x \sin \frac 1x = 0}$$
Therefore, $g$ is differentiable at $x = 0, g'(0) = 0.$
