Explanation

(a)
Since, $y = \frac{3at^{2}}{1 + t^{2}}\ \ and\ \ \ x = \frac{3at}{1 + t^{3}}$
On differentiating given curves w.r.t. $t\$respectively
$\frac{dy}{dt} = \frac{\left( 1 + t^{3} \right)(6at) - 3at^{2}(3t^{2})}{\left( 1 + t^{3} \right)^{2}} = \frac{6at - 3at^{4}}{\left( 1 + t^{3} \right)^{2}}$
and $\frac{dx}{dt} = \frac{\left( 1 + t^{3} \right)(3a) - 3at(3t^{2})}{\left( 1 + t^{3} \right)^{2}} = \frac{3a - 6at^{3}}{\left( 1 + t^{3} \right)^{2}}$
$\therefore\ \frac{dy}{dx}\$=$\ \frac{dy/dt}{dx/dt} = \frac{3at(2 - t^{3})}{3a(1 - 2t^{3})}$=$\frac{t(2 - t^{3})}{(1 - 2t^{3})}$