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
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$

Assuming continuity in the area interval $ (0 < x < 1 )$ and $ (0 > y > -1  ) $ would create problems as $x$ and $y$ are not always real in these areas.
Nay, union of inside parabola areas of $ y_1 = - x^2 $  and $  y_2 = \sqrt{x} $ would violate $ x + y = 0, $ which is only the common chord of intersection of  $ y_1,y_2$. So the shown line joining $(0,0)$ to $(1,-1)$ does not exist as real.
E.g., $(\frac12,  -\frac12)$ does not lie on the
 common line.

