6.236. $\left\{\begin{array}{l}\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}-y^{2}}=6, \\ x y^{2}=6 \sqrt{10} .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ x^{2}-y^{2} \geq 0 .\end{array}\right.$

We square the first equation of the system, then

$$
\begin{aligned}
& x^{2}+y^{2}+2 \sqrt{x^{4}-y^{4}}+x^{2}-y^{2}=36 \Leftrightarrow \sqrt{x^{4}-y^{4}}=18-x^{2} \Rightarrow \\
\Rightarrow & x^{4}-y^{4}=324-36 x^{2}+x^{4}, y^{4}-36 x^{2}+324=0 .
\end{aligned}
$$

From the second equation of the system we have

$$
x=\frac{6 \sqrt{10}}{y^{2}} \Rightarrow y^{4}-36 \cdot \frac{360}{y^{4}}+324=0, y^{8}+324 y^{4}-12960=0
$$

where $y \neq 0 \Rightarrow y^{4}=36, y_{1}=-\sqrt{6}, y_{2}=\sqrt{6}$. Then $x_{1,2}=\frac{6 \sqrt{10}}{6}=\sqrt{10}$.

Answer: $(\sqrt{10} ;-\sqrt{6}),(\sqrt{10} ; \sqrt{6})$.