<s>[INST] 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
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
 [/INST] Assume that $x<0$ and $y>0$, your statement can be written (even if we change $x$ to $-x$),

If $x,y$ are strictly positive such that $(\sqrt{y^2+x}+x)(\sqrt{x^2+y}-y)=y$ then $x=y$.

This equality becomes, $$\overbrace{\dfrac{x+\sqrt{x+y^2}}{y+\sqrt{y+x^2}}}^{A}=\overbrace{\dfrac{y}{x^2+y-y^2}}^{B}.$$ Let us start with $x>y$, clearly $B<1$. Moreover $x\mapsto(x-\sqrt{y+x^2})$ is increasing and $x\mapsto \sqrt{y+x^2}$ strictly increasing, we conclude that $x\mapsto(x+\sqrt{x+y^2})-(y+\sqrt{y+x^2})$ is strictly increasing and thus $A>0$ when $x>y$. We deduces that $A>1$ and therefore $A$ cannot be equal to $B$.
The case $x<y$ can be treated in the same way.
</s>