Locally Lipschitz does not imply $C^1$

The Euclidean norm $f(x)=\|x\|$ is probably the simplest example: the Lipschitz condition comes straight from the triangle inequality, and the non-differentiability is seen from the fact that the restriction of $f$ to the first coordinate axis is $|x_1|$.

By Rademacher's theorem any Lipschitz continuous function is differentiable a.e.

The derivative of a lipschitz function is a bounded measurable function, but this derivative need not be continuous.

Hence the locally lipschitz function need not be $C^1$.

For example $f(x) = \frac {x^i x^j}{ |x|} $ on the unit $n$-ball.