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given the function $f(x)=x^1/3$ defined for any real number $x$ gives us:

This is a function, which means that for any given $x$ value, it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test); and this function also has a vertical tangent at $x=0$.

My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?

*************EDIT***************************

As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?

No, we don't need two vertical points. By the same idea, if the graph of a function $f$ has an horizontal tangent line somewhere, that there has to be two points of the graph of $f$ with the same $y$ coordinate. However, the tangnt at $0$ of $xmapsto x^3$ is horizontal, in spite of the fact that no two points of its graph have the same $y$ coordinate.

Vertical tangent does not mean vertical graph. The function is still one- to -one. If you find its inverse function you see the inverse is one to one and has derivative equal zero so your function is also one to one even if tangent line is vertical.

The tangent line is simply an ideal picture of what you would expect to see if you zoom in around the point $p$.

That being said, the vertical tangent line to the graph $y = sqrt[3]x$ at $(0,0)$ says nothing more than that the graph would look steeper as you further zoom in around $(0, 0)$.

We can also learn several things from this geometric intuition.

1. The line is never required to pass through two distinct points. In fact, in many classical examples such as conic sections, tangent lines pass through a single point.



2. 
   

   It is not essential for $f$ to be differentiable in order to discuss a tangent line to the graph of $f$, although it is a sufficient condition. In your example, differentiating the function $x mapsto sqrt[3]x$ fails to detect the tangent line at $(0,0)$ since it is not differentiable at this point. On the other hand, it perfectly makes sense to discuss vertical tangent line of the curve

   
   
   

   $$ mathcalC = (x, sqrt[3]x) :x in mathbbR = (y^3, y) : y in mathbbR , $$

   
   
   

   and indeed the line $x = 0$ is the tangent line of $mathcalC$ at $(0, 0)$.

   
   

To get a vertical tangent we need 2 vertical points...

Herein lies the error in your assumptions. A tangent intersects a curve at the point of tangency in just one point.

As you observed, $fracdydx$ is not defined at $(x,y) = (0,0).$
And there is no straight segment anywhere on the graph where the slope could be exactly measured by taking any two distinct points on that segment.

But those are not the only ways to define the slope of a curve.

One approach is to take a secant line between $(0,0)$ and a nearby point, and find the limit of the slope of that line as the distance goes to zero.
You can also take the angle between the $x$ axis and the secant line between two points on either side of $(0,0),$ taking a limit as the distance between those points goes to zero.
If the limit is defined (it is not for every curve) then it gives you the tangent line.

Another approach is to parameterize the curve: $x = u(t)$ and $y = v(t)$ where we let $t$ range over some (possibly infinite) interval of $mathbb R.$
Consider the vector $left(fracddtu(t), fracddtv(t)right).$
As long as that vector is not $(0,0)$ we can consider it to be in the same direction as the slope of the curve.
Now let $x = u(t) = t^3,$ $y = v(t) = t.$
Then the vector along the curve at $(x,y) = (0,0)$ is $(0,1),$ which points straight up, so the tangent line is vertical.