Are positive real numbers $x,y$ allowed to be taken out during this proof? 
Prove $$\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.

Can I simplify $\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$
and then $(x^2 - y^2)(x-y) \ge 0$ cancelling out the $xy$?
Is this valid because then 
\begin{align}
(x+y)(x-y)(x-y)&\ge  0\cdot(x-y)^2\ge 0
\end{align}
 which is always true and then prove backwards from here?

If $x,y$ are positive, then $xy>0$; so, when you have
$$\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$$
you're allowed to multiply by $xy$ (and the $\ge $ won't flip), to get
$$(x^2 - y^2)(x-y) \ge 0$$
And from that indeed you can get
$$(x^2 - y^2)(x-y)=(x+y)(x-y)(x-y)=(x+y)(x-y)^2 \ge 0$$
which is true, since $x+y>0$ and $(x-y)^2\ge 0$.
