= Solution
At zero, $|f(x)-f(0)|=|x|\to0$, so the <function> is <continuous>. At a nonzero $x_0$, approach along rational and irrational <sequences>; <density of the rational numbers> and <density of the irrational numbers> give limiting values $x_0$ and $-x_0$, which are different. Thus the \b[<continuity> set is $\boxed{\{0\}}$].
At zero, the <difference quotient> is one for rational nonzero increments and minus one for irrational increments, so the <derivative> does not exist. Elsewhere <differentiability implies continuity>, and continuity already fails. Thus the \b[<differentiability> set is $\boxed{\varnothing}$].
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