Answer
The inequality is valid for values less than -1 and values between 0 and 1 (including them) i.e. $(-\infty,-1]\cap [0,1]$.
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$\dfrac{(x-1)(x+1)}{x}=0$
$\dfrac{(x-1)(x+1)}{x}\cdot x=0\cdot x$
$(x-1)(x+1)=0$
Note that even though we have eliminated $x$, the equation has the restriction $x\ne0$; therefore we are going to count it as a critical point.
$x_1=-1$
$x_2=0$
$x_3=1$
These are the critical points. We are going to take four values: one less than -1; one between -1 and 0; one between 0 and 1; and one more than 1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$\dfrac{(-2-1)(-2+1)}{-2}\leq0$
$\dfrac{(-3)(-1)}{-2}\leq0$
$-1.5\leq0 \rightarrow \text{ TRUE}$
Second test with a value between -1 and 0:
$\dfrac{(-0.5-1)(-0.5+1)}{-0.5}\leq0$
$\dfrac{(-1.5)(0.5)}{-0.5}\leq0$
$1.5\leq0 \rightarrow \text{ FALSE}$
Third test with a value between 0 and 1:
$\dfrac{(0.5-1)(0.5+1)}{0.5}\leq0$
$\dfrac{(-0.5)(1.5)}{0.5}\leq0$
$-1.5\leq0 \rightarrow \text{ TRUE}$
Fourth test with a value more than 1:
$\dfrac{(2-1)(2+1)}{2}\leq0$
$\dfrac{(1)(3)}{2}\leq0$
$1.5\leq0 \rightarrow \text{ FALSE}$
These tests show that the inequality $\dfrac{(x-1)(x+1)}{x}\leq0$ is valid for values less than -1 and values between 0 and 1 (including them) i.e. $(-\infty,-1]\cap [0,1]$
