Let $x,y$ be real numbers such that $$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$ Show that $x+y=0$.
My try: Let $$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$ and then $$\begin{cases} y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\ x^2=b^2+(2b-1)y+y^2\cdots\cdots \end{cases}$$ $(1)+(2)$ then $$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$ so $$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$ we take $(3)$ in $(2)$,we have $$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so $$(2ab-a+b)=0$$ or $$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$ if $$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$ and if $$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$ I don't prove $$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
This is not a solution, but brute force can be used to remove the radicals. Let $A=y^2-x$ and $B=x^2+y$. We have
$$\sqrt{AB}-y\sqrt{A}-x\sqrt{B}+xy=y$$
Isolating $\sqrt{AB}$ and squaring both sides:
$$\sqrt{AB}=y\sqrt{A}+x\sqrt{B}+y(1-x)\quad(1)$$ $$AB=y^2A+x^2B+y^2(1-x)^2+2xy\sqrt{AB}+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
(1) allows us to remove $\sqrt{AB}$. We do this and also recall what $A$ and $B$ equal.
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy\left(y\sqrt{A}+x\sqrt{B}+y(1-x)\right)+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
Group $\sqrt{A}$ and $\sqrt{B}$ terms, then rearrange a bit:
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy^2(1-x)+2y^2\sqrt{A}+2xy\sqrt{B}$$ $$x^2y^2+y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2xy^2+x^2y^2+2xy^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$ $$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$
(1) allows us to sub out the quantity in parentheses:
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y(x-1)+\sqrt{AB}\right)$$ $$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y^2(x-1)+2y\sqrt{AB}$$ $$y^3-x^3-xy=y^4+xy^2+x^4+x^2y-y^2-2x^2y^2+2y\sqrt{AB}$$ $$y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2=2y\sqrt{AB}$$
Squaring both sides, we've reached a goal of no longer having radicals.
$$(y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2)^2=4y^2(y^2-x)(x^2+y)$$
I had a CAS expand this, move it all to one side, and then, as expected, $(x+y)$ factors out of it (twice).
$$(x+y)^2 p(x,y)=0$$
where $$p(x,y)=x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2$$
is a monster. It would be sufficient to show that $p(x,y)$ is never $0$ in the region of the plane where both $\sqrt{A}$ and $\sqrt{B}$ are defined aside from points along $x+y=0$ (like $(0,0)$). This is a pretty messy polynomial, but at least it's a polynomial.
EDIT: This approach seems to be useless; a CAS plot of the zero set of $p$ has several components, all of which are in the region where $\sqrt{A}$ and $\sqrt{B}$ are defined. They must be extraneous solutions from the squaring that was done twice.