It's not what you think!

First, factor the equation -

$\frac { (x+3)(x-3) }{ (x+4)(x-4) } \ge 0$

Now for the expression to have a value more than or equal to $0$ , it must be positive or 0. So now we need to find the values of $x$ which when substituted in the expression make it positive or 0.

We do that by determining the critical numbers which can be found by setting the numerator and denominator equal to zero.

So for numerator the critical numbers are

$(x+3)(x-3)=0$

$x=3$ or $x=-3$

And for denominator, critical numbers are

$(x+4)(x-4)=0$

$x=4$ or $x=-4$

So the critical numbers of the inequality are $x = 3, x = -3, x = 4, x = -4$

Now there are many ways to solve it but I think the easiest way is to use a number line. I will not be able to show the exact thing here but I will give it a go.

Arrange the critical numbers on a number line and find out the range of $x$ .

That table shows the value of expression (whether positive or negative) given on the left when $x$ is assigned different values.

Since we want the whole expression to be positive, we can either take the value of $x$ between $-3$ and $3$ , greater than $4$ or lesser than $-4$ (Note that negative multiplied by negative is positive).

So here is the range formed from above data -

$x < -4$ or $x > 4$ or $-3\le x\le 3$

Note that $x$ cannot be equal to $4$ or $-4$ as that will make the denominator $0$ which will make the fraction indeterminate.

I know that I'm not really good at explaining things, so if you have any query about my answer, then please ask in the comments section below!

First of all we need to identify how we can we satisfy this inequality. There are 3 possibilities.

$1\cdot$ $x^2-9=0$ this will give value of inequality is 0.

$2\cdot$ both numerator and denominator is positive means $x^2-9 \gt 0$ and $x^2-16\gt0$ .

$3\cdot$ and last one is both numerator and denominator is negative means $x^2-9 \lt 0$ and $x^2-16\lt 0$

so first constraint give $x=3,-3$

second one give $x\gt 3$ and $x\gt 4$ after combined give $x\gt 4$ so as $x\lt -4$ . (numeric value of x must greater than 4 and in negative numbers the smaller the number give bigger numeric value ).

third constraint give $x\lt 3$ and $x\lt 4$ after combine them give $x\lt 3$ so as $x\gt -3$ .

so we have range of x is $x\lt-4,-3\le x\le3,x\gt4$

$x^2\ge 9~~\implies ~x\ge ~3 ~and~x\le~-3. ~~~~But ~x\neq \pm 4.\\So~~ \color{#D61F06}{x <-4 , -3\le x\le 3, x > 4.}$

For the numerator:

$x^2-9 = (x-3)(x+3)$

For the denominator:

$x^2-16 = (x-4)(x+4)$

Note that $x^2-16 \ne 0$ because division by zero is undefined. So the solutions, $x$ , between -3 and 3 are inclusive, but the solutions greater than 4 or less than -4 are exclusive.

This gives you the roots as: $(x-3)(x+3) \ge 0 \\ => x = -3, 3 \\ (x-4)(x+4) \gt 0 \\ => x = -4, 4$

The simplest way to solve this problem is to choose test points in between all the roots ${-4, -3, 3, 4}$ . Personally you are really only testing whether or not any of these are an inclusive range (e.g. $-3 \ge x \ge 3$ ), so what I would do initially is test only two points: $x = -5, 0, \frac{7}{2}$ . The reason for this is because -5 is less than -4, 0 is between -3 and 3, and 3.5 is between 3 and 4. The solution at 0 will tell me the behavior of $x$ between -3 and 3 as well as -4 and 4; -5 will tell me the behavior when $x$ is less than -4 and also when it is greater than 4; 3.5 will tell me the behavior when $x$ is between either -4 and -3 or 3 and 4 (in the latter two cases this is because this is an even function and thus symmetric about the origin).

Using these two values shows that:

$x = 0: \\ \frac{x^2-9}{x^2-16} = \frac{9}{16} \\ \\ x = -5: \\ \frac{x^2-9}{x^2-16} = \frac{16}{9} \\ \\ x = \frac{7}{2}: \\ \frac{x^2-9}{x^2-16} = -\frac{13}{15}$

This means that the range -3, 3 is inclusive, the values greater than 4 or less than -4 are included, but the values between -4..-3 and 3..4 are not in the solution set.

So you have:

$x \lt -4, -3 \le x \le 3, x \gt 4$