For $x<0$, $|x|=-x$ so $f(x)=\dfrac{x}{-x}=-1$ (constant) ⇒ continuous. For $x>0$, $f(x)=-1$ ⇒ continuous. At $x=0$: $\displaystyle \lim_{x\to0^-} f(x)=-1,\quad \lim_{x\to0^+} f(x)=-1,\quad f(0)=-1$ Hence $f$ is continuous at $x=0$. Therefore, $f$ is continuous for all real $x$.