Use sequential criterion to show $f$ is continuous exactly at where $\sin\vert x \vert\equiv0 $ .
Added : Let $x \neq k \pi, k \in \Bbb{Z}$
If $\{x_n\} \in \Bbb{Q}$ such that $x_n \rightarrow x$, then by continuity of sine function, $$\lim f(x_n)=\lim \sin \vert x_n \vert=\sin (\lim \vert x_n \vert )=\sin \vert x \vert \neq 0$$
Whereas if $\{x_n\} \in \Bbb{Q}^c$ such that $x_n \rightarrow x$, then $$\lim f(x_n)=0$$
so $f$ is discontinuous at $x \neq k \pi, k \in \Bbb{Z}$
Hint :
Use sequential criterion to show $f$ is continuous exactly at where $\sin\vert x \vert\equiv0 $ .
Added : Let $x \neq k \pi, k \in \Bbb{Z}$
If $\{x_n\} \in \Bbb{Q}$ such that $x_n \rightarrow x$, then by continuity of sine function, $$\lim f(x_n)=\lim \sin \vert x_n \vert=\sin (\lim \vert x_n \vert )=\sin \vert x \vert \neq 0$$ Whereas if $\{x_n\} \in \Bbb{Q}^c$ such that $x_n \rightarrow x$, then $$\lim f(x_n)=0$$ so $f$ is discontinuous at $x \neq k \pi, k \in \Bbb{Z}$