The identity $x/y-y/x = (y+x)(y^{-1}-x^{-1}).$ Difference of perfect square asserts that the expression $y^2-x^2$ factorizes as $(y+x)(y-x)$. On the train home last night, I noticed a variant on this. Namely, that $x/y-y/x$ factorizes as $(y+x)(y^{-1}-x^{-1}).$ Explicitly:

Proposition. Let $R$ denote a ring, and consider $x,y \in R$ such that both $x$ and $y$ are units. Then the following identity holds. $$\frac{x}{y}-\frac{y}{x} = (y+x)(y^{-1}-x^{-1})$$

For example, in the rational numbers, we have:
$$\frac{3}{2}-\frac{2}{3} = 5\cdot \left(\frac{1}{2}-\frac{1}{3}\right)$$

Questions.
Q0. Does this variant on "difference of perfect squares" have a name?
Q1. Is it part of a larger family of identities that includes $y^2-x^2 = (y-x)(y+x)$, and other variants?
Q2. Does it have any interesting uses or applications, e.g. in calculus?

If anyone can think of better tags, please retag.

Not as far as I know, but notice that your identity is
$$ (x+y)\frac{x-y}{xy} = (xy)^{-1}(x+y)(x-y) = \frac{x^2-y^2}{xy} = \frac{x}{y}-\frac{y}{x}, $$
so I'm not sure I'd say that it's really distinct from difference-of-two-squares.
As far as generalisation goes, you've got the old
$$ x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dotsb+ xy^{n-2}+xy^{n-1}). $$
