Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$ Whilst solving a problem, the following system of non-linear equations arose:
$$0= \frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}$$
$$0= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}$$
I have tried to solve it via brute force and by trying to add them both in a way that could give me the answer, but I still haven't got it.
I can't manage to find the intersection of the two parametric equations that these equations give. I know that I can find approximate solutions with the Wolfram Alpha tool, but I would like to know, if any, an analytical or strong numerical way to proceed.
Note: Out of all the solutions ($8$ it seems, $4$ real and $4$ complex), I only need the one that takes real values for $x$ and $y$, with $0<x<y<1$.

Provided that $x\ne0;\;x\ne 1;\;y\ne0;\;y\ne 1;\;x\ne y$ we simplify the denominators and get
$$
\begin{cases}
-x^4+4 x^2 y-2 x y^2-2 x y+y^2=0\\
-2 x^2 y+x^2+4 x y^2-2 x y+y^4-4 y^3+2 y^2=0\\
\end{cases}\tag{1}
$$
add the two equation in $(1)$
$$-x^4+4 x^2 y-2 x^2 y+x^2+4 x y^2-2 x y^2-2 x y-2 x y+y^4-4 y^3+2 y^2+y^2=0$$
which can be factored
$$(y-x) (x+y-1) \left(x^2+x+y^2-3 y\right)$$
First factor $y-x=0$ is discarded for the conditions above
$x+y-1=0\to y=1-x$ give
$$x^4+6 x^3-11 x^2+6 x-1=0$$
two real solutions
$$(x=-7.5619;\;y=8.5619);\;(x=0.31926;\;y=0.68074)$$
$x^2+y^2+x-3 y=0$
I can't see elementary methods to get the other solutions. A graph can help, though.

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