Explanation

(c)
$\because\ \ \ y = \left( \frac{c^{6} - a^{2}x^{4}}{b^{2}} \right)^{\frac{1}{4}}$
Let $f(x) = xy = \left( \frac{c^{6}x^{4} - a^{2}x^{8}}{b^{2}} \right)^{\frac{1}{4}}$
$\Rightarrow \ \ \ f^{'}(x) = \frac{1}{4}\left( \frac{c^{6}x^{4} - a^{2}x^{8}}{b^{2}} \right)^{- 3/4}\left( \frac{4x^{3}c^{6}}{b^{2}} - \frac{8x^{7}a^{2}}{b^{2}} \right)$
Put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ \ x = \pm \frac{c^{3/2}}{2^{1/4}\sqrt{a}}$
$\therefore\ \ \ \ \ \ f\left( \frac{c^{3/2}}{c^{1/4}\sqrt{a}} \right) = \frac{c^{3}\ }{\sqrt{2ab}}$