Explanation

(d)
Given, $x^{2} + y^{2} = t - \frac{1}{t}\ and\ x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}$
$\Rightarrow \ \ x^{4} + y^{4} + 2x^{2}y^{2} = t^{2} + \frac{1}{t^{2}} - 2$
$\Rightarrow \ \ x^{4} + y^{4} + {2x}^{2}y^{2} = x^{4} + y^{4} - 2\$
$\Rightarrow \ \ x^{2}y^{2} + 1 = 0 \Rightarrow y^{2} = \frac{- 1}{x^{2}}\$
On differentiating w.r.t. $x,\ we\ get\$
$2y\ \frac{dy}{dx} = \frac{2}{x^{3}} \Rightarrow \frac{dy}{dx} = \frac{1}{x^{3}y}$