Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$

Take the cue of $0<x<y<1$ and observe that $x+y=1$ satisfies both equations $$\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$$ which leads to $\frac{1}{x^2}-\frac{1}{(1-2x)^2}-\frac{1}{(1-x)^2}=0$, or $$x^4+6x^3-11x^2+6x -1=0$$ Factorize $$[x^2+3(1+\sqrt2)x-(1+\sqrt2)][x^2+3(1-\sqrt2)x-(1-\sqrt2)]=0$$ to obtain the solution in the domain $(0,1)$

$$x= \frac12\left( \sqrt{31+22\sqrt2}-3\sqrt2-3\right) $$ $$ y= \frac12\left(- \sqrt{31+22\sqrt2}+3\sqrt2+5\right)$$

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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