Cone of tangents of $S=\{(x,y):y\ge-x^3\}$ Find the cone of tangents for the set $S$ at the point $\overline x=(0\ 0)^t$.
$$S=\{(x,y):y\ge-x^3\}$$
The cone of tangents of $S$ at $\overline x$ is the set of directions $d$ such that $d=\lim_{k\to\infty}\lambda_k(x_k-\overline x),\ \lambda_k>0,x_k\in S$ for all $k$ and $x_k\to\overline x$.
In this particular case, $d=\lim_{k\to\infty}\lambda_k(x,y)$
what should I do next?
I've searched online but I only find 'solved' exercises that just write the answer and does not says how the cone was found with details.
I greatly appreciate any assistance you may provide.  
 A: I think that your definition of cone of tangents needs to be cleaned up a little. Try this:

The cone of tangents of a set $S$ at a point $\overline{x}$ is the set of directions $d$ such that there exists a sequence of scalars $\{\lambda_k\}$ and vectors $\{x_k\}$ for $k=1,2,\dots,$ such that $\lambda_k>0$ for all $k$, $x_k\in{S}$ for all $k$, $x_k\to\overline{x}$ and $d=\lim_{k\to\infty}\lambda_k(x_k-\overline{x})$.

The easiest choice for the scalars $\lambda_k$ is usually to take $x_k\neq\overline{x}$ for all $k$, and define $\lambda_k=1/\|x_k-\overline{x}\|$ for all $k$. 
In this case, if you look at a picture of the set $S$, it seems like you should be able to move in any direction $d$ from $(0,0)$ such that $d_2\geqslant0$. Indeed, this is correct:
$$ D_0=\{d\in\mathbb{R}^2\ |\ d_2\geqslant0\}. $$
I won't completely prove this, but to give you a flavor, let's prove that $(-1,0)\in{D_0}$. The sequence we want to construct will snake along the boundary of $S$: let's define
$$ x_k=\begin{pmatrix}-1/k\\1/k^3\end{pmatrix}$$
for all $k\in\{1,2,\dots\}$. This sequence, for $k=1,\dots,20$, looks like this:

It's easy to see that $x_k\in{S}$ for all $k$, as required. Now, per the suggestion above, let's define
$$ \lambda_k=\frac{1}{\|x_k-\overline{x}\|_\infty}=\frac{1}{\|x_k\|_\infty}=k$$
for all $k$. I chose the infinity norm $\|\cdot\|_\infty$ to make the resulting proof easier. This would work with any norm (indeed, since all norms on $\mathbb{R}^n$ are equivalent). Clearly $\lambda_k>0$ for all $k$, again as required. It's also clear that
$$ \lim_{k\to\infty}x_k=\begin{pmatrix}0\\0\end{pmatrix} $$
as required. All we need to check now is that
$$
\lim_{k\to\infty}\lambda_k(x_k-\overline{x})=\lim_{k\to\infty}k\begin{pmatrix}-1/k\\1/k^3\end{pmatrix}=\lim_{k\to\infty}\begin{pmatrix}-1\\1/k^2\end{pmatrix}=\begin{pmatrix}-1\\0\end{pmatrix}
$$
as required.
A: There are several first-order cones that appear in context of KKT conditions and constraint qualifications (like feasible directions, attainable directions etc), but all those cones, including the tangent cone, are between two standard cones for the linearized problem that are easy to calculate (here $I$ is the set of indices for active constraints):
\begin{align}
G_0&=\{d\colon \nabla g_i(\bar x)^Td<0,\,i\in I\},\\
G'&=\{d\colon \nabla g_i(\bar x)^Td\le 0,\,i\in I\}.
\end{align}
In our case, $g(x,y)=-x^3-y\le 0$ is active at the origin, so
\begin{align}
G_0&=\{d\colon (0\ \ {-}1)\,d<0\}=\{d\colon d_2>0\},\\
G'&=\{d\colon (0\ \ {-}1)\,d\le 0\}=\{d\colon d_2\ge 0\}.
\end{align}
Since
$$
\text{closure}\,G_0\subset T\subset G'
$$
and $\text{closure}\,G_0=G'$ (the closed upper half-plane), the tangent cone $T$ is also the closed upper half-plane.
P.S. The condition $\text{closure}\,G_0=G'$ is known as Cottle's constraint qualification (which is equivalent to the Mangasarian-Fromovitz constraint qualification in case of no equality constraint).
