Simplifying fractions with fractions I am trying to simplify 
$$\frac{\frac{y}{x} - \frac{x}{y}}{\frac{1}{y} - \frac{1}{x}}$$
I make the top part into 
$$\frac{y^2 - x ^2}{xy}$$
I know the bottom can be rewritten to just be multiplied into the top as its inverse.
$$\frac{y^2 - x ^2}{xy} * (y - x)$$
$$\frac{y - x }{xy} $$
This of course is wrong, but I do not know why, it appears to be correct to me. Nothing I did is mathematically wrong, it follows all the rules. What went wrong?
 A: The inverse of the denominator
$$\frac{1}{y}-\frac{1}{x}$$
is not $y-x$. The denominator is equal to
$$\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy},$$
so its inverse is
$$\frac{xy}{x-y},$$
so the entire fraction is
$$\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}=\frac{\frac{y^2-x^2}{xy}}{\frac{1}{y}-\frac{1}{x}}=\frac{\frac{y^2-x^2}{xy}}{\frac{x-y}{xy}}=\frac{y^2-x^2}{xy}\cdot\frac{xy}{x-y}=\frac{y^2-x^2}{x-y}$$
There's one more simplification you can make at this step; do you see it?
A: No, you didn’t follow the rules: there is nothing in the rules that lets you transform
$$\frac1{\frac1y-\frac1x}$$
into $y-x$, which is what you’ve done when you go from
$$\frac{\frac{y^2-x^2}{xy}}{\frac1y-\frac1x}$$
to 
$$\frac{y^2-x^2}{xy}\cdot(y-x)\;.$$
You can’t invert the denominator until you’ve transformed it into a single fraction:
$$\frac1y-\frac1x=\frac{x-y}{xy}\;.$$
Now you can invert and multiply:
$$\frac{\frac{y^2-x^2}{xy}}{\frac1y-\frac1x}=\frac{y^2-x^2}{xy}\cdot\frac{xy}{x-y}=\frac{y^2-x^2}{x-y}=\frac{(y-x)(y+x)}{x-y}=-(y+x)\;.$$
A: $$\frac{\frac{y^2-x^2}{xy}}{\frac{x-y}{xy}}=\frac{y^2-x^2}{x-y}=\frac{(y-x)(y+x)}{-(y-x)}=\frac{y+x}{-1}=-y-x$$
