Vertical tangent at the origin in the Folium of Descartes So, for the folium of Descartes, there is both a horizontal tangent, and a vertical tangent, at $(0,0)$. 
Using parametric equation, 
$$x = \frac{3t}{1+t^3} \qquad y= \frac{3t^2}{1+t^3}$$
For horizontal tangent, I get it. But what about vertical tangent? 
There exists vertical tangent, when $\frac{dx}{dt} = 0$, but $\frac{dx}{dt}$ equals to $0$ only when $t = (\frac{1}{2})^{0.333}$. This gives another point where vertical tangent is present. 
Then, how can I get by calculation that there is a vertical tangent at $(0,0)$ using these parametric equations?
 A: The parameterization has the unusual property that the double-point at the origin is reached when $t=0$ and when $t=\pm \infty$. (The asymptotic part corresponds to behavior near $t=-1$.)
The horizontal tangent corresponds to how the point moves near $t=0$; that's easy enough.
However, as $t$ heads towards infinity, both $x$ and $y$ change very little. This is observed in the derivatives
$$\frac{dx}{dt}=\frac{3\left(1-2t^3\right)}{\left(1+t^3\right)^2} \qquad \frac{dy}{dt}=\frac{3t\left(2+t^3\right)}{\left(1-t^3\right)^2}$$
that both approach $0$ for large $t$. Yet, it wouldn't be appropriate to say that the curve as both a horizontal and vertical tangent at $t=\infty$. We resolve the ambiguity by combining the parametric derivatives into the proper $xy$-derivative
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t(2+t^3)}{1-2t^3}$$
In this form, we see that $t=\pm \infty$ corresponds to an infinite slope (or, if you prefer, it corresponds to a reciprocal-slope of zero), hence, a vertical tangent line.
So, this is something to keep in the back of your mind about parameterized curves. After all, "slope" is a relationship between $x$ and $y$, not a relationship between $x$ or $y$ and some third value. While it is often reasonable to consider coordinate-wise derivatives separately (as in determining the horizontal tangents here), sometimes that separation obscures exactly what you need to see.
A: $$ x'(t) = \frac{3(1-2t^3)}{(1+t^3)^2} ,\ y'(t) = \frac{3t(2-t^4)}{(
1+t^3)^2} $$
Hence $\frac{dy}{dt} = \frac{y'}{x'} = \frac{t(2-t^4)}{1-2t^3}$
which goes to $\infty$ when $t\rightarrow \infty$.
