The given function is $f(x)=|x|-|x+1|$.The two functions, $g$ and $h$, are defined as$g(x)=|x|$ and $h(x)=|x+1|$Then, $f=g-h$The continuity of $g$ and $h$ is examined first.$g(x)=|x|$ can be written as$g(x)=\begin{cases} -x, & \text{if } x<0 \\ x, & \text{if } x\ge 0 \end{cases}.$Clearly, $g$ is defined for all real numbers.Let $c$ be a real number.Case IIf $c<0$, then $g(c)=-c$ and $\lim\limits_{\overline{n}\to c}g(x)=\lim\limits_{x\to c}(-x)=-c$ $\therefore \lim\limits_{x\to c}g(x)=g(c)$Therefore, $g$ is a continuous at all points $x$, such that $x < O$ Case IIIf $c>$, then $g(c)=c$ and $\lim\limits_{x\to c}g(x)=\lim\limits_{x\to c}x=c$$\therefore \lim\limits_{x\to c}g(x)=g(c)$Therefore, $g$ is continuous at all points $x$, such that $x>0$Case IIIIf $c=o$, then $g(c)=g(o)=o$$\lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}(-x)=0$$\lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}(x)=0$$\therefore \lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}(x)=g(0)$Therefore, $g$ is continuous at $x=o$From the above three observation, it can be concluded that $g$ is continuous at all points.$h(x)=|x+1|$ can be written as$h(x)=\begin{cases} -(x+1), & \text{if } c<-1 \\ x+1, & \text{if } x\ge -1 \end{cases}.$Clearly, $h$ is defined for every real number.Let c be a real number.Case I :If $c<-1$, then $h(c)=-(c+1)$ and $\lim\limits_{\overline{n}\to c}h(x)=\lim\limits_{x\to c}[-(x+1)]=-(c+1)$$\therefore \lim\limits_{h\to c}h(x)=h(c)$Therefore, $h$ is continuous at all points $x$, such that $x<-1$Case II:If $c>-1$, then $h(c)=c+1$ and $\lim\limits_{x\to c}h(x)=\lim\limits_{x\to c}(x+1)=c+1$$\therefore \lim\limits_{n\to c}h(x)=h(c)$Therefore, $h$ is continuous at all points $x$ such that $x>-1$.Case IIIIf $c=-1$, then $h(c)=h(-1)=-1+1=0$$\lim\limits_{x\to -1}h(x)=\lim\limits_{x\to -1}[-(x+1)]=-(-1+1)=0$$\lim\limits_{x\to -1}h(x)=\lim\limits_{x\to -1}(x+1)=(-1+1)=0$$\therefore \lim\limits_{x\to -1}h(x)=\lim\limits_{x\to -1}h(x)=h(-1)$Therefore, $h$ is continuous at $x=1$From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.$g$ and $h$ are continuous functions. Therefore, $f=g h$ is also a continuous function.Therefore, f has no point of discontinuity.