Answer
$$
f(x)=\left\{\begin{array}{ll}{\frac{3}{x-1},} & {x \neq 1} \\ {3,} & {x=1}\end{array}\right.
$$
value $x=1$ , at which $f $ is not continuous
Work Step by Step
$$
f(x)=\left\{\begin{array}{ll}{\frac{3}{x-1},} & {x \neq 1} \\ {3,} & {x=1}\end{array}\right.
$$
We obtain that
\begin{equation}
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{3}{x-1}= \infty
\end{equation}
\begin{equation}
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} \frac{3}{x-1}=-\infty
\end{equation}
$$
\lim _{x \rightarrow 1^{-}} f(x) \ne \lim _{x \rightarrow 1^{-}} f(x)
$$
therefore $$ \lim _{x \rightarrow 1} f(x) $$ does not exist
therefore at $x=1$ , $f $ is not continuous
