Im my high-school calculus class I was recently taught that for a parametric equation (with $x$ and $y$ in terms of $t$, for convenience), a vertical tangent occurs only when $\frac{dx}{dt}=0$ and $\frac{dy}{dt} \neq 0$. I've found this is unanimously supported online as well; for example, here: http://mathonline.wikidot.com/horizontal-and-vertical-tangents-of-parametric-curves.
My problem with this is that it seems to be too restrictive. The underlying logic behind setting $\frac{dx}{dt} = 0$ is that $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, so if $\frac{dx}{dt}$ = 0, $\frac{dy}{dx}$ will approach $\pm \infty$ (assuming a nonzero numerator) so there will be a vertical tangent.
However, by the same logic above, $\frac{dy}{dt}$ could just approach $\pm \infty$ directly, and there would be a vertical tangent (or an asymptote). For example, if $y = \sqrt[3]{t}$, and $x = (t - 1)^2$, the traditional method would be to set $\frac{dx}{dt} = 2(t-1)=0 \rightarrow t=1$, so $((1-1)^2, \sqrt[3]{1}) = (0,1)$ would have a vertical tangent (the only one). But setting $\frac{dy}{dt} = \frac{1}{3 \sqrt[3]{t^2}} = \pm \infty \rightarrow t=0$, so at $((0-1)^2, \sqrt[3]{0}) = (1,0)$ there should also be a vertical tangent, and graphing confirms there is (not complete): 
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Technically the graph is not a function, but restricting the domain of $t$ should solve that if it's even relevant. I would assume that I'm doing something wrong here, since I haven't but able to find a single source that did the above. But the graph seems to make it patently obvious that there are two points of vertical tangency, not one.
Have I done something incorrect? If not, why are vertical tangents taught like this?