Answer
$$
f(x)=\frac{-5+x}{3x(3x+1)}
$$
$ x$-values where the given function is discontinuous. are
$$x=0 , \quad x=-\frac{1}{3} $$
and
$$f(0), \quad f(-\frac{1}{3})$$
do not exist, also
$$ \lim _{x \rightarrow { -\frac{1}{3}}} f(x) ,\quad \lim _{x \rightarrow 0} f(x)
$$
are not exist
Work Step by Step
$$
f(x)=\frac{-5+x}{3x(3x+1)}
$$
This rational function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=0 $ and $x=-\frac{1}{3} $ .
so the given function is discontinuous at $x=0 $ and $x=-\frac{1}{3} $, therefore $f(0), f(-\frac{1}{3})$ do not exist.
We investigate $ \lim _{x \rightarrow 0} f(x) $ at $x=0$ first. From the left, where $x$-values are less than $0$,
$$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{-5+x}{3x(3x+1)} \right)=\infty
$$
From the right, where $x$-values are greater than 0,
$$
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{-5+x}{3x(3x+1)} \right)=-\infty
$$
so, $ \lim _{x \rightarrow 0} f(x) $ does not exist since
$$
\lim _{x \rightarrow 0^{+}} f(x) \ne \lim _{x \rightarrow 0^{-}}f(x)
$$
Now, we investigate $ \lim _{x \rightarrow -\frac{1}{3}} f(x) $ at $x=-\frac{1}{3}$ first. From the left, where $x$-values are less than $-\frac{1}{3}$,
$$
\lim _{x \rightarrow{ -\frac{1}{3}}^{-}} f(x)=\lim _{x \rightarrow { -\frac{1}{3}}^{-}}\left(\frac{-5+x}{3x(3x+1)} \right)=-\infty
$$
From the right, where $x$-values are greater than $ -\frac{1}{3}$,
$$
\lim _{x \rightarrow { -\frac{1}{3}}^{+}} f(x)=\lim _{x \rightarrow { -\frac{1}{3}}^{+}}\left(\frac{-5+x}{3x(3x+1)} \right)=\infty
$$
so, $ \lim _{x \rightarrow { -\frac{1}{3}}} f(x) $ does not exist since
$$
\lim _{x \rightarrow { -\frac{1}{3}}^{+}} f(x) \ne \lim _{x \rightarrow { -\frac{1}{3}}^{-}}f(x)
$$
Hence, $ x$-values where the function is discontinuous. are
$x=0 $ and $x=-\frac{1}{3} $ .
$f(0), f(-\frac{1}{3})$ do not exist, also
$ \lim _{x \rightarrow { -\frac{1}{3}}} f(x) $, $ \lim _{x \rightarrow 0} f(x) $ do not exist.