In which cyclic cubic number fields does there exist this type of unit? Let $K$ be a cyclic cubic number field with conductor $f$ and ring of integers $\mathcal{O}_K$. 
Define $K$ to be blue if and only if  $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in K$}.$$
Define $K$ to be green if and only if $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in \mathcal{O}_K$}.$$ (So green implies blue).
Question 1: Are all cyclic cubic number fields blue?
Question 2: What is the density of green number fields restricted to blue number fields? That is, defining $$B_N:=\{K:K\text{ is a blue (cyclic cubic) number field of conductor }<N\},$$ $$G_N:=\{K:K\text{ is a green (cyclic cubic) number field of conductor }<N\},$$ what is 
$$
\lim_{N\to\infty} \frac{\#G_N}{\#B_N}?
$$
(and does the limit exist?)
Question 3:  Define $$\mathcal{G}:=\{f: K \text{ is green, where $K$ is a cyclic cubic number field of conductor $f$}\}.$$ What is $\mathcal{G}$ explicitly?
Remarks: I wrote some magma code that proved that $K$ is blue for all of the 1822 cubic cyclic number fields given from LMFDB (http://www.lmfdb.org/NumberField/start=0&degree=3&galois_group=C3&count=20). The code also explicitly gives the minimal polynomial of $w$. Here are the first few examples.
\begin{align*}
 f=7,  \quad &  t^3 - 2t^2 - t + 1
 \\
 f=9,  \quad &  t^3 - 3t + 1
 \\
 f=13,  \quad &  t^3 + t^2 - 4t + 1
 \\
 f=19,  \quad &  t^3 - 5t^2 + 2t + 1
 \\
 f=31,  \quad &  t^3 - (5/2)t^2 - (1/2)t + 1
\end{align*}
The polynomials above prove that $\{7,9,13,19\}\subseteq\mathcal{G}$. Notice that for $f=31$, this polynomial implies $K$ of conductor $31$ is blue, but it may or may not be green.
 A: One can consider an asymptotic version of question 2. It is well-known, by the work of Delone and Faddeev (though supposedly this result was due to Levi) that cubic orders are parametrized by $\text{GL}_2(\mathbb{Z})$-equivalence classes of integral binary cubic forms. Since maximal, irreducible cubic orders are in one-to-one correspondence with cubic fields, it suffices to consider those cubic forms that correspond to maximal irreducible cubic orders with square discriminant. By this paper of Bhargava and Shnidman (https://projecteuclid.org/euclid.ant/1513730135) we know that all binary cubic forms corresponding to cubic orders of this type take the shape
$$\displaystyle F(x,y) = ax^3 + bx^2 y + (b-3a)xy^2 - ay^3, a, b \in \mathbb{Z}.$$
Moreover, those $F$ with bounded discriminant lie in the ellipse defined by 
$$\displaystyle b^2 - 3ab + 9a^2 \leq X.$$
Now, we wish to describe the norm form of the corresponding cubic order $\mathcal{O}_F$ in terms of $F$, or rather, in terms of $a,b$. I can't seem to find a reference (I was told that this was in Bhargava's thesis but it doesn't seem to be in there), but one can write down the norm form for $\mathcal{O}_F$ explicitly. Here is the idea: the norm form for the cubic order is the same as the ideal norm form for the trivial class in the ideal class group, and by Bhargava's Higher Composition Laws II, the class is parametrized by the pair of symmetric matrices
$$\displaystyle (A,B) = \left(\begin{pmatrix} 0 & 0 & 1 \\ 0 & -a & 0 \\ 1 & 0 & 3a - b \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 1 & b & 0 \\ 0 & 0 & -a \end{pmatrix}\right).$$
Using a construction found here https://www.springer.com/gp/book/9781441909985 (the Petitjean chapter), one can construct an auxiliary ternary quadratic form from $A,B$, and using these three quadratic forms, construct a cubic form $G$. One can then compute explicitly that $G$ is proportional to its Hessian, hence it is a decomposable form. I was told that Bhargava proved at some point that $G$ is in fact the norm form for $\mathcal{O}_F$, and when $\mathcal{O}_F$ is a maximal irreducible cubic order, also the norm form for the corresponding cubic field. An explicit form for $G$ is
$$\displaystyle G(x,y,z) = x^3 - a^3 y^3 - a^3 z^3 - a(b-3a)^2y^2 z - 2a^2 by^2 z - ab^2 yz^2 + 2a^2(b-3a)yz^2 + a(b-3a)xy^2 - b(b-3a)xyz - 3a^2 xyz +  bx^2 y - abxz^2 - (b-3a)x^2 z.$$ 
The question is then asking for simultaneous solutions to the equation $G(x,y,z) = -1, G(1-x, -y, -z) = -1$ (the norm form comes from a normalized basis of the shape $\langle 1, \omega, \theta \rangle$). This boils down to first finding an integral point on an affine quadric in $x,y,z$, and whether such a conic contains an integral point should give some information on the existence of $N_{K/\mathbb{Q}}(w) = N_{K/\mathbb{Q}}(1 - w) = -1$. 
A: A supplementary answer which sheds a little more light on question 3, and which extends another result of M.-N. Gras from 1973:
The "green" fields among cyclic cubic fields $K$ are precisely those which contain a Galois-invariant monogenic order $\mathbb{Z}[y] \subseteq \mathcal{O}_K$.
The easy direction: When $-w$ is a root of a Shanks polynomial $t^3-at^2-(a+3)t-1$, then we can take $y=\pm w$, optionally translated by a rational integer.
The other direction is not much harder. When $y^\sigma \in \mathbb{Z}[y]$, say $y^\sigma=a_0+a_1y+a_2y^2$ for rational integer $a_0,a_1,a_2$, then
$$y^{\sigma^2}-y^\sigma = a_1(y^\sigma-y) +a_2(y^\sigma-y)^2$$
lies in the principal ideal $(y^\sigma-y)$. Applying the automorphism $\sigma$ repeatedly, we find
$$(y^{\sigma^2}-y^\sigma) \subseteq (y^\sigma-y) \subseteq (y-y^{\sigma^2}) \subseteq (y^{\sigma^2}-y^\sigma)$$
so this ideal is invariant under $\sigma$. Therefore the three summands in
$$\frac{y^\sigma-y}{y^\sigma-y} + \frac{y^{\sigma^2}-y^\sigma}{y^\sigma-y} + \frac{y-y^{\sigma^2}}{y^\sigma-y} = 0$$
are algebraic units (of positive norm). The second and third summands will thus be roots of Shanks polynomials with integer parameter, and we can pick e.g.
$$w=\frac{y^{\sigma^2}-y}{y^\sigma-y}.$$
The maximal order $\mathcal{O}_K$ is always Galois-invariant, but not necessarily monogenic; when it isn't, the field could still contain a non-maximal Galois-invariant order and thus exceptional units. In these cases, the discriminant $(a^2+3a+9)^2$ of the Shanks polynomial will necessarily be a proper multiple of $f^2$.
A: Thank you @GNiklasch for answering question 1. I had the following realisation towards answering questions 3 and 2...
As in
Ennola, Veikko; Turunen, Reino, On cyclic cubic fields, Math. Comput. 45, 585-589 (1985). ZBL0582.12002., we can write the conductor $f$ as
$$
f=\frac{a^2+3b^2}{4} 
$$
for integers $a$ and $b$ such that
\begin{align*}
    &a \equiv 6 \bmod 9 \text{ and } b\equiv 3 \text{ or } 6 \bmod 9 \text{ where } b>0 
    & \text{ if } 3|f,\\
    &a \equiv 2 \bmod 3 \text{ and } b\equiv 0 \bmod 3 \text{ where } b>0 
    & \text{ if } 3\nmid f.\\
\end{align*}
Now define 
$$
c:= \left\{ 
\def\arraystretch{1.5}
\begin{array}{l l}
    \frac{-3}{2}\left(1 + \frac{a}{b}\right)  & \text{if } 3| f \\
    \frac{-3}{2}\left(1 - \frac{a}{b}\right) & \text{if } 3\nmid f.
\end{array}
\right.
$$
If $c\in \mathbb{Z}$ then $K$ is green and the minimal polynomial of $\omega$ is given by 
$$
m_\omega(x):= x^3 + cx^2 - (c+3)x + 1.
$$
