Differentiable but not continuously differentiable

1. For $x\ne0$, by the product and chain rules,
$$f'(x)=2x\sin\frac1x-\cos\frac1x.$$
At $0$, the difference quotient is
$$\frac{f(h)-f(0)}h=h\sin\frac1h,$$
and $|h\sin\frac1h|\le|h|\to0$. So $f'(0)=0$ and $f$ is differentiable everywhere.

2. As $x\to0$, the term $2x\sin\frac1x\to0$, but $\cos\frac1x$ has no limit: it equals $1$ along $x_n=\frac1{2n\pi}$ and $-1$ along $y_n=\frac1{(2n+1)\pi}$. So $f'(x)$ has no limit at $0$, and in particular $f'$ is not continuous at $0$, although $f'(0)$ exists.

Remark. A derivative can be discontinuous. It cannot however have a jump discontinuity (see the problem on the limit of the derivative): its discontinuities are always of this oscillating kind.