Explanation

(a)
Given, $2f\left( x^{2} \right) + 3f\left( \frac{1}{x^{2}} \right) = x^{2} - 1$ ...(i)
Replacing $x$ by $\frac{1}{x}$, we get
$2f\left( \frac{1}{x^{2}} \right) + 3f\left( x^{2} \right) = \frac{1}{x^{2}} - 1$ ...(ii)
On multiplying Eq. (i) by 2, Eq. (ii) by 3 and subtracting Eq. (i) from Eq. (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5f\left( x^{2} \right) = \frac{3}{x^{2}} - 1 - 2x^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f\left( x^{2} \right) = \frac{1}{5x^{2}}(3 - x^{2} - 2x^{4})$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f\left( x^{4} \right) = \frac{1}{5x^{4}}(3 - x^{4} - 2x^{8})$ [Replacing $x$ by $x^{2}$]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\left( 1 - x^{4} \right)(2x^{4} + 3)}{5x^{4}}$