= Solution
Away from zero, the product and composition of the displayed smooth <functions> are <continuous> and <differentiable>. At zero,
$$
|f(x)|\le|x|\longrightarrow0,
$$
so $f$ remains <continuous>. Hence the \b[<continuity> set is $\boxed{\mathbb R}$].
The <difference quotient> at zero is $f(h)/h=\sin(1/h)$. For $h_n=(\pi/2+2\pi n)^{-1}$ it equals one, and for $k_n=(3\pi/2+2\pi n)^{-1}$ it equals minus one. Both <sequences> tend to zero, so no <derivative> exists there. For $x\ne0$, the <product rule> and <chain rule> give
$$
\boxed{f'(x)=\sin(1/x)-\frac{\cos(1/x)}x.}
$$
Thus the \b[<differentiability> set is $\boxed{\mathbb R\setminus\{0\}}$].
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