$2.343 A=\frac{\sqrt{4(x-\sqrt{y})+y x^{-1}} \cdot \sqrt{9 x^{2}+6 \sqrt[3]{2 y x^{3}}+\sqrt[3]{4 y^{2}}}}{6 x^{2}+2 \sqrt[3]{2 y x^{3}}-3 \sqrt{y x^{2}}-\sqrt[6]{4 y^{5}}}$.

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$2.343 A=\frac{\sqrt{4(x-\sqrt{y})+y x^{-1}} \cdot \sqrt{9 x^{2}+6 \sqrt[3]{2 y x^{3}}+\sqrt[3]{4 y^{2}}}}{6 x^{2}+2 \sqrt[3]{2 y x^{3}}-3 \sqrt{y x^{2}}-\sqrt[6]{4 y^{5}}}$.

The above text is already in English, so no translation is needed.
Solution. Obviously, $y \geq 0$. Then $x>0$, otherwise $4(x-\sqrt{y})+y x^{-1}\sqrt{y}, \\ A=\frac{2 x-\sqrt{y}}{\sqrt{x}(2 x-\sqrt{y})}\end{array} \Leftrightarrow\left\{\begin{array}{l}4 x^{2}>y \geq 0, \\ A=\frac{1}{\sqrt{x}} .\end{array}\right.\right.$

Answer: if $x>0, y>4 x^{2}$, then $A=-\frac{1}{\sqrt{x}}$; if $x>0,0 \leq y<4 x^{2}$, then $A=\frac{1}{\sqrt{x}}$.