$\text{ The point(s) on the curve }{y}^3+3x^2=12y\text{ where the tangent is vertical is (are)}$

Differentiating the given curve w.r.t.  x, we get

$3y^2\frac{dy}{dx}+6x=12\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-\frac{2x}{y^2-4}$

$\text{ At point where the tangent(s) is (are) vertical, }\frac{dy}{dx}\text{ is not }$defined,  i.e. at those points.

$\begin{array}{l}{y}^2-4=0\Rightarrow y=\pm 2\\ \text{ When }y=2,8+3x^2=12\left(2\right)\Rightarrow 3x^2=16\Rightarrow x=\pm \frac{4}{\sqrt{3}}\end{array}$

$\text{ when }y=-2,-8+3x^2=-24\Rightarrow 3x^2=-16\text{. This is not possible}$

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