Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$ 
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
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$

A more generalised approach over my earlier post. This is not intended to be an exhaustive proof but an experimental one. Constructive comments are most welcome. 
Let 
$$\sqrt{y^2-x}-x=Ay^n\qquad \cdots (1)\\
\sqrt{x^2+y}-y=\frac {y^{1-n}}A \qquad \cdots (2)\\$$
such that the original equation $$\left(\sqrt{y^2-x}-x\right)\left(\sqrt{x^2+y}-y\right)=y$$
is satisfied as required.
From $(1)$,
$$\begin{align}
\sqrt{y^2-x}&=x+Ay^n\\
y^2-x&=x^2++2Axy^n+A^2y^{2n}\\
y^{2n}A^2+2xy^nA+(x^2-y^2+x)&=0\\
A^2+\frac {2x}{y^n}A+\frac{(x^2-y^2+x)}{y^{2n}}&=0\qquad \qquad \qquad \qquad \cdots (3)
\end{align}$$
From $(2)$,
$$\begin{align}
\sqrt{x^2+y}&=y+\frac {y^{1-n}}A\\
A\sqrt{x^2+y}&=Ay+y^{1-n}\\
A^2(x^2+y)&=A^2y^2+2Ay^{2-n}+y^{2(1-n)}\\
(x^2-y^2+y)A^2-2y^{2-n}A-y^{2(1-n)}&=0\\
A^2-\frac{2y^{2-n}}{x^2-y^2+y}A-\frac{y^{2(1-n)}}{x^2-y^2+y}&=0\qquad \qquad \qquad \cdots (4)
\end{align}$$
Equating coefficients of $A^1$:
$$\begin{align}\frac{2x}{y^n}&=-\frac{2y^{2-n}}{x^2-y^2+y}\\
y^2&=-x(x^2-y^2+y)\qquad \qquad \qquad \qquad \qquad \qquad \cdots (5)\end{align}$$
Equating coefficients of $A^0$:
$$\begin{align}\frac{x^2-y^2+x}{y^{2n}}&=-\frac{y^{2(1-n)}}{x^2-y^2+y}\\
y^2&=-(x^2-y^2+x)(x^2-y^2+y)\qquad \cdots (6)\end{align}$$
(5)=(6):
$$\begin{align}x(x^2-y^2+y)&=(x^2-y^2+x)(x^2-y^2+y)\\
(x^2-y^2)(x^2+y^2-y)&=0\\
(x-y)(x+y)(x^2-y^2+y)&=0\\
\Rightarrow x-y=&0, x+y=0, x^2-y^2+y=0\end{align}$$
Checking by substitution into the original equation shows that only 
$$x+y=0$$
is valid.
This graph created on desmos.com might help illustrate the approach:
https://www.desmos.com/calculator/qrlbgbalix
