If you take the reciprocal in an inequality, would it change the > / <...

If a and b have the same nonzero sign, then
$a<b⟺\frac{1}{a}>\frac{1}{b}$
(i.e., taking reciprocals reverses the inequality).
If a and b have opposite (nonzero) signs, then
$a<b⟺\frac{1}{a}<\frac{1}{b}$
(i.e., taking reciprocals preserves the inequality).
These follow from the fact that the function $f(x)=1/x$ defined on the nonzero reals is strictly decreasing and sign-preserving on each component $(-\mathrm{\infty },0)$ and $(0,\mathrm{\infty })$ of its domain.
If either of a or b is zero, then you can't take reciprocals.
Finally, compound inequalities like $a<b<c$ should be separated into "$a<b$ and $b<c$" and each component considered separately.
I also remark that inverting a sum is not the same as inverting the addends separately.

It depends if $x$ and $y$ are the same sign.
Case 1: $0<x<y$ then $0<x(1/y)<y(1/y)$ and $0<x/y<1$ and $0<x/y(1/x)<1(1/x)$ so $0<1/y<1/x$
If both positive, flip.
Case 2: $x<0<y$ then $x/y<0<1$. Then as $x<0$ we flip when we do $x/y\ast (1/x)>0>1\ast (1/x)$ so $1/y>0>1/x$ so $1/x<0<1/y$. Don't flip.
Case 3: $x<y<0$ then $x/y>1>0$ and $1/y<1/x<0$. Flip if they are the same sign.
But FOR THE LOVE OF GOD!!!!!!! the reciprical of $1/x-1/4$ is !!!!!!!NOT!!!!!! $x/1-4/1$!!!!!!!!
It is $\frac{1}{1/x-1/4}=\frac{4x}{4-x}$