Answer
The inequality is valid for values less than -1 and values more than 3 (not including them) i.e. $(-\infty,-1)\cap (3,\infty)$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$\dfrac{x-3}{x+1}=0$
$\dfrac{x-3}{x+1}\cdot(x+1)=0(x+1)$
$x-3=0$
$x=3$
It is important to note that even though we have eliminated the expression $x+1$, the equation has the restriction $x\ne-1$; therefore we are going to count it as a critical point.
$x_1=-1$
$x_2=3$
These are the critical points. We are going to take three values: one less than -1; one between -1 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$\dfrac{-2-3}{-2+1}>0$
$\dfrac{-5}{-1}>0$
$5>0 \rightarrow \text{ TRUE}$
Second test with a value between -1 and 3:
$\dfrac{0-3}{0+1}>0$
$\dfrac{-3}{1}>0$
$-3>0 \rightarrow \text{ FALSE}$
Third test with a value more than 3:
$\dfrac{4-3}{4+1}>0$
$\dfrac{1}{5}>0$
$\dfrac{1}{5}>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x-3}{x+1}>0$ is valid for values less than -1 and values more than 3 (not including them) i.e. $(-\infty,-1)\cap (3,\infty)$