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?

What you did in step 1 amounts to multiply both sides by $(x-4)(x+3)$. Unfortunately, you have to reverse the inequation if this expression is negative, and leave it as is if it is positive. And as you don't know the sign of this product…
You can simplify solving this inequation writing both sides in canonical form:
$$\frac{x-3}{x-4}=1+\frac1{x-4}>\frac{x+4}{x+3}=1+\frac1{x+3}\iff\frac1{x-4}>\frac1{x+3}$$
Multiplying both members by $(x-4)^2(x-3)^2$ (which is positive on the domain of the inequation), we obtain:
$$(x-4)(x+3)^2> (x-4)^2(x+3)\iff7(x-4)(x+3)>0\iff\begin{cases}x<-3\\\text{or}\\x>4\end{cases}$$
