Inequalities with $x^2$ in the denominator. $\frac{81-x^2}{x^2}\le0$ I can handle the $81-x^2\le0$, but not with the $x^2$ in the denominator. I don't think can multiply both sides $x^2$. Help needed here.
 A: Since $x^2 \ge 0$ for all $x \in \mathbb{R}$, but with $x^2$ in the denominator meaning $x \neq 0$ (since with $x = 0$, the numerator would be $81 \neq 0$) so $x^2 \gt 0$, you can multiply by $x^2$ without changing the inequality, i.e., you will get
$$81 - x^2 \le 0 \tag{1}\label{eq1A}$$
as you stated. Although it's not an issue in this case, don't forget to account for $x \neq 0$ if need be. I trust you can finish the rest yourself.
A: $\frac{A}{B}\le 0\iff \begin{cases}A\le 0\\ B>0\end{cases}\lor\begin{cases}A\ge0\\ B<0\end{cases}$, therefore $$\frac{81-x^2}{x^2}\le 0\iff \begin{cases}81-x^2\le 0\\ x^2>0\end{cases}\lor\begin{cases}81-x^2\ge0\\ x^2<0\end{cases}$$
et cetera.
A: Yes we can multiply by $x^2$ providing that $x^2\neq 0$ to obtain
$$\frac{81-x^2}{x^2}\le0 \iff 81-x^2\le0 \iff x^2\ge81 \iff x\in(-\infty,-9]\cup[9,\infty)$$
since $x^2>0$ doesn't change the sign of the inequality.
As a concrete example let consider consider
$$\frac{81-100}{100}\le0 \iff 100\cdot \frac{81-100}{100}\le 100\cdot 0 \iff 81-100 \le 0$$
A: In homogenous inequalities with fractions like this you don't have to multiply, but factorize: 
$$\frac{81-x^2}{x^2}\le0 \Rightarrow \frac{(9-x)(9+x)}{x^2}\le0$$
And finding the zero and nonzero points: $x=-9,x=9,x\ne 0$, check the signs of the intervals:
$$x\in (-\infty,-9], \frac{(+)\cdot (-)}{(+)}\le 0 \ \ \ \color{green}{\checkmark}\\
x\in (-9,0), \frac{(+)\cdot (+)}{(+)}\not\le 0 \ \ \ \  \color{red}{\emptyset} \\
x\in (0,9), \frac{(+)\cdot (+)}{(+)}\not\le 0 \ \ \ \ \color{red}{\emptyset}\\
x\in [9,\infty), \frac{(-)\cdot (+)}{(+)}\le 0 \ \ \ \color{green}{\checkmark}$$
Alternatively:
$$\frac{81-x^2}{x^2}\le0 \Rightarrow \frac{81}{x^2}-1\le 0 \Rightarrow \frac{81}{x^2}\le 1 \Rightarrow \frac{1}{x^2}\le \frac1{81} \Rightarrow |x|>9.$$
