How to find the coordinates of each point on the curve where the tangent line is vertical?

Explanation:
We evaluate the derivative of the function at the point of tangency to find $m=$ the slope of the tangent line at that point.
$m=0$ means the tangent line is horizontal at that point
$m=\pm \infty$ means the tangent line is vertical at that point.

$\frac{dy}{dx}=\frac{3y-2x}{6y-3x}=\pm \infty$

$6y-3x=0$

$6y=3x$
$x=2y$

We plug this into the function to solve for one coordinate of the point:
$\begin{array}{l}{x}^2+3y^2=1+3xy\\ 4y^2+3y^2=1+6y^2\\ y^2=1\\ y=\pm 1\\ y=1,\therefore x=2y=2\\ y=-1,\therefore x=2y=-2\end{array}$
$(2,1)$ and $(-2,-1)$ are the points where the function has vertical tangents
The graphs of the function and its vertical tangents are shown below: