If $a^2{x}^4+b^2{y}^4=c^6$, then the maximum value of xy is

Given $a^2{x}^4+b^2{y}^4=C^6\to \left(1\right)$
AM ≥ GM
$\frac{a^2{x}^4+b^2{y}^4}{2}\ge \sqrt{\left(a^2{x}^4\right)\left(b^2{y}^4\right)}$

$\begin{array}{l}\frac{a^2{x}^4+b^2{y}^4}{2}\ge ab{x}^2{y}^2\\ a^2{x}^4+b^2{y}^6\ge 2ab{x}^2{y}^2\\ c^6\ge 2ab{x}^2{y}^2\\ \frac{c^6}{2ab}\ge x^2{y}^2\\ \Rightarrow x^2{y}^2\le \frac{c^6}{2ab}\\ xy\le \sqrt{\frac{c^6}{2ab}}\\ xy\le \frac{c^3}{\sqrt{2ab}}\end{array}$