If $f(x) = |\sin x|,$ then

We have, $f(x) = |\sin x|$

Let $f(x) = {\mathop{\rm vou}\nolimits} (x) = v[u(x)]$ [where, $u(x) = \sin x$ and $v(x) = |x|$]
$= v(\sin x) = |\sin x|$

where, $u(x)$ and $v(x)$ are both continuous.

Hence, $f(x) =$ $vo\,\,u(x)$ is also a continuous function but $v(x)$ is not differentiable at $x = 0$.

So, $f(x)$ is not differentiable where $\sin x = 0 \Rightarrow x = n\pi ,n \in Z$

Hence, $f(x)$ is continuous everywhere but not differentiable at $x = n\pi ,n \in Z$.