Solution:
The provided function is
$f(x)= \begin{cases}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 \\ 6 x+2, & \text { if } x \geq 3\end{cases}$
Given here the $f(x)$ is defined for $x \leq-3$ or $(-\infty,-3)$, for $-3<x<3$ and even for $x \geq 3$ or $(3, \infty)$.
As a result,
Domain of $f$ is $(-\infty,-3) \cup(-3,3) \cup(3, \infty)=(-\infty, \infty)=R$
For all $x<-3, f(x)=|x|+3=-x+3$ is a polynomial and therefore continuous and for all $x(-3<x<3), f(x)=-2 x$ is a continuous and a continuous function and even for all $x>3, f(x)=6 x+2$.
As a result, $\mathrm{f}(\mathrm{x})$ is continuous on $\mathrm{R}-\{-3,3\}$.
And, of domain $R$, $x=-3$ and $x=3$ are partitioning points.
So now, Left Hand limit =
$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 5}(|x|+3)=\lim _{x \rightarrow 5}(-x+3)=3+3=6$
Right Hand limit $\lim _{x \rightarrow 3+} f(x)=\lim _{x \rightarrow 3^{-}}(-2 x)=(-2)(-3)=6$
And $f(-3)=|-3|+3=3+3=6$
As a result, $f(x)$ is continuous at $x=-3$.
Now again the, Left Hand limit =
$=\lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 3}(-2 x)=-2(3)=-6$
Right Hand limit =
$\lim _{x \rightarrow 3+} f(x)=\lim _{x \rightarrow 3^{-}}(6 x+2)=6(3)+2=20$
Since, $\lim _{x \rightarrow 3} f(x) \neq \lim _{x \rightarrow 3} f(x)$
As a result, $\lim _{x \rightarrow 3} f(x)$ does not exist and therefore $f(x)$ is discontinuous at only $x=3$.