#### Answer

$$
f(x)=\frac{3}{x}+\frac{x-1}{x^{2}-1}
$$
values of $x$, at which $f $ is not continuous are $x = 0$, $x = 1$ and $ x = -1$

#### Work Step by Step

$$
f(x)=\frac{3}{x}+\frac{x-1}{x^{2}-1}
$$
suppose
$$
f(x)=g(x)+h(x)
$$
such that
$$g(x)= \frac{3}{x}, \quad \quad h(x)=\frac{x-1}{x^{2}-1}$$
The function $ g(x)=\frac{3}{x}$ being graphed is a rational function, and hence is continuous at every number except $ x=0$. Now consider the other function
$$
x^{2}-1=0
$$
yields discontinuities at $x = 1$ and at $ x = -1$.
Therefore the function
$$
h(x)=\frac{x-1}{x^{2}-1}
$$
is continuous for all real numbers $x$ except $x = 1$ , $ x = -1$
Thus
$$
f(x)=\frac{3}{x}+\frac{x-1}{x^{2}-1}
$$
is continuous for all real numbers $x$ except $x = 0 $, $x = 1$ and $ x = -1$.