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?


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...

As others have pointed out, this is the crux of the misunderstanding. That said, I'd like to try and succinctly highlight the core issue: and that is that derivatives are not defined by the secant of two points, but rather by the limit of secants approaching a certain point.
In the OP example, letting $x = 0$, as we take other domain values $x'$ trending closer to $x$, inspecting the trend of secant slopes is the essential meaning of the derivative there. In this case, the secant slopes grow larger without bound, which is the indication of an undefined (or infinite) derivative, and hence a vertical asymptote. The definition from Wikipedia:

Limit definition

A function $f$ has a vertical tangent at $x = a$ if the difference quotient used to define the derivative has infinite limit:
$$\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = {+\infty}\quad\text{or}\quad\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = {-\infty}.$$

The Wikipedia article on vertical tangents uses the same $f(x) = x^{1/3}$ example, so hopefully that's clarifying. I suspect that this kind of misunderstanding may be the result of a particular course of study failing to emphasize the limit definition of the derivative.
