Gfrktyl

What's the result of:
$$frac{1}{|{x}|}-frac{x^2}{|x|^3}$$
Is it
$$frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|x|}=0$$
or
$$frac{1}{|{x}|}-frac{x^2}{|x|^2x}=frac{1}{|{x}|}-frac{1}{x}frac{x^2}{|x|^2}=frac{1}{|{x}|}-frac{1}{x}=left{begin{matrix}frac{2}{x},x<0\ 0,x>0 end{matrix}right.$$

As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.

The error in the second proof is in the very beginning, namely $frac{1}{|{x}|}-frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.

Your first conclusion is right since $$|x|^3=|x|^2cdot |x|=x^2cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3ne x^3= |x|^2cdot x$$

Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then

$$frac{1}{|{x}|}-frac{x^2}{|x|^3}=frac{1}{|{x}|}-frac{x^2}{|x|x^2}=frac{1}{|{x}|}-frac{1}{|{x}|}=0$$

For the same reason the second one is wrong.

As the other answers already told you, the first one is true.

But maybe this way of thinking gives you an intuition: Both terms $frac{1}{|x|}$ and $frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.