The point(s) on the curve y^3+3 x^2=12 y where the tangent is vertica…

We have,

$y^3+3 x^2=12 y$                .....(i)

$\Rightarrow 3 y^2 \frac{d y}{d x}+6 x=12 \frac{d y}{d x}$          [Diff. w.r.t. x]

$\Rightarrow 3\left(y^2-4\right) \frac{d y}{d x}=-6 x \Rightarrow \frac{d y}{d x}=-\frac{2 x}{y^2-4}$

At point(s) where the tangent(s) is (are) vertical, $\frac{d y}{d x}$ is not defined.

∴  $y^2-4=0 \Rightarrow y= \pm 2$

From (i), we find that

$y=2 \Rightarrow x= \pm \frac{4}{\sqrt{3}}$

and, $y=-2 \Rightarrow x^2=\frac{-16}{3}$, which is not possible.

Hence, the required points are $( \pm 4 / \sqrt{3}, 2)$