 Solution

For $m=0$, continuity fails. Take $x_k=(\pi /2+2\pi k{)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2\pi k{)}^{-1}$. Both tend to zero, but the corresponding function values are $1$ and $-1$. Thus there is no limit, and the function is not differentiable either, since differentiability would imply continuity.

For $m=1$, the bound $|x\sin (1/x)|\le |x|$ and the squeeze theorem give continuity at zero. Its difference quotient is $\sin (1/h)$, however, which has no limit by the same two sequences. It is therefore not differentiable there.

For $m=2$, the bound $|x^2\sin (1/x)|\le x^2$ proves continuity. Its difference quotient is $h\sin (1/h)$, whose absolute value is bounded by $|h|$, so the derivative exists and equals zero. The complete classification is