Explanation

(d)
Given curve is $y^{3} + 3x^{2} = 12y$
On differentiating w.r.t. $y$, we get
$3y^{2}\frac{dy}{dx} + 6x = 12\frac{dy}{dx}$
$\Rightarrow \ \frac{dy}{dx}\left( 3y^{2} - 12 \right) + 6x = 0$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{6x}{12 - 3y^{2}}$
$\Rightarrow \ \frac{dx}{dy} = \frac{12 - 3y^{2}}{6x}$
Since, tangent is parallel to $y$-axis
$\frac{dy}{dx} = 0\ \ \Rightarrow 12 - 3y^{2} = 0$
$\Rightarrow \ \ y^{2} = 4\ \ \ \Rightarrow \ \ \ y = \pm 2$
Then, $x = \pm \frac{4}{\sqrt{3}}$
At $y = - 2,\ x$ cannot be real
$\therefore$ The required point is $\left( \pm \frac{4}{\sqrt{3}},\ 2 \right)$