If $a^2x^4+b^2y^4=c^6$, then the maximum value of $xy$ Simply subsisting the value of $y$ in $xy$ will not work, but I don’t know any other method to solve it. Can I get a hint?

Method 1
You could use the parametric equation of ellipse to solve it.
Let $p=x^2,q=y^2$ then $x^4 = p^2,y^4=y^2$, the equation turns to
$\frac{p^2}{c^6/a^2} + \frac{q^2}{c^6/b^2}=1$, which is the ellipse function. Thus
$p = \frac{c^3}{a} \cos\theta$
$q = \frac{c^3}{b} \sin\theta,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$
Since $xy = \sqrt{pq}$, $\max xy = \sqrt{\frac{c^6}{2ab}}$
the maximum is taken when $\theta=2k\pi+ \frac{\pi}{2},k\in\mathbb{Z}$
Method 2
Use the AM-GM inequality:
$c^6 = (ax^2)^2 + (by^2)^2 \ge 2ab(xy)^2$
thus
$xy \le \sqrt{\frac{c^6}{2ab}}$
equal is taken when $a x^2 = b y^2$.
