<s>[INST] For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side 
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point. 
 [/INST] Alt. hint:   following up on OP's $^{\tiny\text{(?)}}\,$ approach, with the missing $\,=3\,$ appended...

I tried to expand the left hand side: $$\;xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)} = 3 \tag{1}$$

Before that, multiplying the original equation by $\,\left(x\color{red}{-}\sqrt{x^2+3}\right)\left(y\color{red}{-}\sqrt{y^2+3}\right)\,$ gives:
$$
(-3)\,(-3) = 3 (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) \;\;\iff\;\; (x-\sqrt{x^2+3})(y-\sqrt{y^2+3}) = 3 \tag{2}
$$
Expanding $(2)$ and subtracting from $(1)$ then gives:
$$\require{cancel}
\begin{align}
y\sqrt{x^2+3} + x\sqrt{y^2+3} = 0 \;\;&\iff\;\; y\sqrt{x^2+3} = -x\sqrt{y^2+3} \\
 &\implies\;\;y^2(\cancel{x^2}+3)=x^2(\cancel{y^2}+3) \\
 &\iff\;\; x^2 = y^2 \\
 &\iff\;\; (x-y)(x+y) = 0
\end{align}
$$
The case $\,x-y = 0\,$ has the only solution $\,x=y=0\,$, which leaves $\,x+y=0\,$ either way.
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