How can a "proper" function have a vertical slope? Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:

Since $f$ is a function, 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).
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?
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?

The tangent line is simply an ideal picture of what you would expect to see if you zoom in around the point.
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Hence, the vertical tangent line to the graph $y = \sqrt[3]{x}$ at $(0,0)$ says nothing more than that the graph would look steeper and steeper as we zoom in further 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, as the idea of a tangent line itself does not impose such an extraneous condition.
For instance, tangent lines pass through a single point even in many classical examples such as conic sections. On the other extreme, a tangent line can pass through infinitely many points of the original curve as in the example of the graph $y = \sin x$.
2. Tangent line is purely a geometric notion, hence it should not depend on the coordinate system being used.
On the contrary, identifying the curve as the graph of some function $f$ and differentiating it does depend on the coordinates system. In particular, it is not essential for $f$ to be differentiable in order to discuss a tangent line to the graph $y = f(x)$, although it is a sufficient condition.
OP's example serves as a perfect showcase of this. Differentiating the function $f(x) = \sqrt[3]{x}$ 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 the vertical tangent line to the curve
$$ \mathcal{C} = \{(x, \sqrt[3]{x}) :x \in \mathbb{R} \} = \{(y^3, y) : y \in \mathbb{R} \}, $$
and indeed the line $x = 0$ is the tangent line to $\mathcal{C}$ at $(0, 0)$.
