If ${\mathrm{a}}^2{\mathrm{x}}^4+{\mathrm{b}}^2{\mathrm{y}}^4={\mathrm{c}}^6 (\mathrm{a},\mathrm{b},\mathrm{x},\mathrm{y},\mathrm{c}>0)$, then the maximum value of $xy$ is

${\mathrm{a}}^2{\mathrm{x}}^4+{\mathrm{b}}^2{\mathrm{y}}^4={\mathrm{c}}^6$
or      $\mathrm{y}={\left(\frac{{\mathrm{c}}^6-{\mathrm{a}}^2{\mathrm{x}}^4}{{\mathrm{b}}^2}\right)}^{1/4}$
or     $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xy}=\mathrm{x}{\left(\frac{{\mathrm{c}}^6-{\mathrm{a}}^2{\mathrm{x}}^4}{{\mathrm{b}}^2}\right)}^{1/4}$
             $={\left(\frac{{\mathrm{c}}^6{\mathrm{x}}^4-{\mathrm{a}}^2{\mathrm{x}}^8}{{\mathrm{b}}^2}\right)}^{1/4}$
Differentiate f (x) w.r.t x, we get
${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{{\mathrm{c}}^6{\mathrm{x}}^4-{\mathrm{a}}^2{\mathrm{x}}^8}{{\mathrm{b}}^2}\right)}^{-3/4}\cdot \left(\frac{4{\mathrm{x}}^3{\mathrm{c}}^6}{{\mathrm{b}}^2}-\frac{8{\mathrm{x}}^7{\mathrm{a}}^2}{{\mathrm{b}}^2}\right)$
${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=0\Rightarrow \frac{4{\mathrm{x}}^3{\mathrm{c}}^6}{{\mathrm{b}}^2}-\frac{8{\mathrm{x}}^7{\mathrm{a}}^2}{{\mathrm{b}}^2}=0$
or    ${\mathrm{x}}^4=\frac{{\mathrm{c}}^6}{2{\mathrm{a}}^2}\text{ or }\mathrm{x}=\pm \frac{{\mathrm{c}}^{3/2}}{2^{1/4}\sqrt{\mathrm{a}}}$
At   $\mathrm{x}=\frac{{\mathrm{c}}^{3/2}}{2^{1/4}\sqrt{\mathrm{a}}},\mathrm{f}\left(\mathrm{x}\right)$will be maximum. So,
      $\mathrm{f}\left(\frac{{\mathrm{c}}^{3/2}}{2^{1/4}\sqrt{\mathrm{a}}}\right)={\left(\frac{{\mathrm{c}}^{12}}{2{\mathrm{a}}^2{\mathrm{b}}^2}-\frac{{\mathrm{c}}^{12}}{4{\mathrm{a}}^2{\mathrm{b}}^2}\right)}^{1/4}={\left(\frac{{\mathrm{c}}^{12}}{4{\mathrm{a}}^2{\mathrm{b}}^2}\right)}^{1/4}$
                                                                 $=\frac{{\mathrm{c}}^3}{\sqrt{2\mathrm{ab}}}$
Alternative method:
Since $AM\ge GM$.,
      $\frac{{\mathrm{a}}^2{\mathrm{x}}^4+{\mathrm{b}}^2{\mathrm{y}}^4}{2}\ge \sqrt{{\mathrm{a}}^2{\mathrm{x}}^4{\mathrm{b}}^2{\mathrm{y}}^4}$
or    ${\mathrm{abx}}^2{\mathrm{y}}^2\le \frac{{\mathrm{c}}^6}{2}$
or    $\mathrm{xy}\le \frac{{\mathrm{c}}^3}{\sqrt{2\mathrm{ab}}}$
Hence, maximum value of $xy$ is $\frac{{\mathrm{c}}^3}{\sqrt{2\mathrm{ab}}}$.