 Solution

Write the symmetric difference quotient as the average of two ordinary difference quotients:If $f$ is differentiable at $x$, both terms tend to the derivative $f^{\prime }(x)$. Hence the symmetric derivative satisfies $\boxed{f^s(x)=f^{\prime }(x)}$.

The absolute value function $f(x)=|x|$ is continuous, and $f^s(0)=0$ because $|h|=|-h|$. Its right one-sided derivative at zero is $1$ and its left one-sided derivative is $-1$, so it is not differentiable at zero. Away from zero it is differentiable, hence has a symmetric derivative there as well. This supplies an example meeting the hypotheses on, for example, $(-1,1)$.