Inequation: quadratic difference equations Given:
$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$
Step 1:
$$(x + 3)(x - 3) > (x + 4)(x - 4)$$
Step2 : Solving step 1:
$$x^2 - 3^2 > x^2 - 4^2$$
*Step 3: 
$ 0 > -16 + 9$ ???
As you see, I can delete the $x^2$, but there is no point in doing that.
What should be the next step?

Where could the inequality reverse? Candidates are the two singularities (i) $x=-3$ and $x=4$ and (ii) the places where $(x+3)(x-3)=(x+4)(x-4)$. By your calculation there are no such places.
Now use three test points, one to the left of $x=-3$, one between $x=-3$ and $x=4$, and one to the right of $x=4$.
For example, take the test point $-4$ to the left of $-3$. It is clear that the left-hand side is greater than the right-hand side, so the inequality holds in the interval $(-\infty,-3)$. Now use two other test points to finish.
