Geometric interpretation of the norm $\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}$ 
Let $p:\mathbb R^2 \to \mathbb R$ be a norm so that
  $$
\|\vec x\|
={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}
={{\|\vec x\|_1\over 3}}+{2\|\vec x\|_\infty\over 3}.
$$
  I need to graph the neighbourhood of radius $1$ around $(0,0)$: $V_1 ((0,0))$ with this norm, but I don't even know the points that are in this neighbourhood I really don't know how to geometrically visualize it .

I tried to separate the norm in to parts: I want that to find all $(x_1, x_2) \in \mathbb{R}^2$ that satisfy
$$
{(|x_{1}|+|x_{2}|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}
< 1
$$
so:
$$
\frac{|x_1|+|x_2|}{3}
< \frac{1}{2}
\qquad \text{and} \qquad
\frac{2\max(|x_1|,|x_2|)}{3} < {1\over 2}.
$$
I know that the first inequality is a rotated square (geometrically) and the second one is a square, but from this point I don't see how to find the points that satisfy the given norm and visualize it geometrically.
 A: Note the expression for the norm can be altered to;
$$ ||(x_1, x_2)|| = \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} $$
Hence the graph you look for is points $(x_1, x_2)$ such that, 
$$  \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} \lt 1 \iff 3 \max \{|x_1|, |x_2|\} + \min \{|x_1|, |x_2|\}\lt 3 $$
I'm not sure about this part. Needs verification: 
From here I think you must map the following regions on the $(x, y)$ plane in the corresponding regions. 
For $ |y| \ge |x| $; we get the equation to be $ 3|y| + |x| \lt 1 $
$$ 3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \ge 0  $$
$$ 3 y - x \lt 3 \;\; \text{for } \;\; x \lt 0, y \ge 0  $$
$$ - 3 y -  x \lt 3 \;\; \text{for } \;\; x \lt 0, y \lt 0  $$
$$ - 3 y + x \lt 3 \;\; \text{for } \;\; x \ge 0, y \lt 0  $$
Now I think we need to similarly map $ 3|y| + |x| \lt 1 $ for $ |x| \ge |y| $.
Note however that the $8$ separate equations we get are confined to $8$ disjoint regions in the plane. The $8$ sub-quadrants or octants if you will. So I'm guessing it can be done.  
A: Using $\max(a,b) = \frac{|a| + |b| + |a-b|}{2}$ we can rewrite your norm as
$$
\| (x,y) \|
:= \frac{|x| + |y]}{3}+ \frac{|x| + |y| + ||x| - |y||}{3}
$$
Now we want to solve
\begin{align}
\frac{|x| + |y|}{3}+ \frac{|x| + |y| + ||x| - |y||}{3} < 1
\iff & \frac{2}{3}\big(|x| + |y|\big) + \frac{1}{3} \big||x| - |y|\big| < 1 \\
\iff & 4 | x | + 4 | y | + 2 \big||x| - |y|\big| < 6 \qquad (\ddagger)
\end{align}
We now have to distinguish some cases:
Case 1: $x,y \ge 0$.
Then we have
$$
(\ddagger) \iff 4x + 4y + 2 | x - y | < 6 \qquad (\star)
$$
Case 1.1: $x \ge y \ge 0$.
Then we have
$$
(\star)
\iff 4x + 4y + 2x - 2y < 6
\iff 6x + 2y < 6
\iff 3x + y < 3.
$$
Case 1.2: $y \ge x \ge 0$.
Then we have
$$
(\star)
\iff 4x + 4y + -2x + 2y < 6
\iff x + 3y < 3
$$
All the other cases can be done analogously, you obtain $-x+3y<3$ and $-3x+y<3$ for the second quadrant (going counter clockwise), $-x-3y<3$ and $-3x-y < 3$ for the third and $x-3y < 3$ and $3x - y < 3$ for the fourth quadrant.
This traces out the following regular octagon:

This is the intersection of two rotated squares.
Its area ($\approx 3.02$) is roughly $\frac{3}{4}$ of the area of the square $[-1,1]^2$.
You can rewrite the octagon as
$$
\big\{ (x,y) \in \mathbb{R}^2:  ax + by < 3, \ a,b \in \{\pm 1, \pm 3\}, |a| \ne |b| \big\}
$$
and I'm not sure if this characterisation can be obtained more easily (with less or without case distinctions) from the above inequality.
Note: Your approach one part < 1/2 and other part < 1/2 gives some but not all of the solutions, do you know, why?
A: Essentially the same as other answers but phrased differently. If $X=(x,y)$,  $\|X\|:=\frac13\|X\|_1 + \frac23\|X\|_\infty$, then we are looking for the (lets say open) ball of radius 1 around $(0,0)$,
$$ B:=V_1((0,0)) = \{X\in\mathbb R^2 : \|X\|<1\}$$
Observe the symmetries
\begin{align}
\|(x,y)\|&= \|(-x,y)\|,\\
\|(x,y)\|&= \|(x,-y)\|,\\
\|(x,y)\|&= \|(y,x)\|,\\
\|(x,y)\|&= \|(-x,-y)\|.\end{align}
These imply that the set $B$ is symmetric across the $y$-axis, the $x$-axis, across the line $y=x$, and across the line $y=-x$ respectively. So it suffices to describe $B$ in the octant $y\ge x\ge 0$, where since $\max(x,y)=y$,
$$ \|X\| < 1 
 \iff \frac13(x+y) + \frac23 y<1$$
i.e. the set is described by 8 copies of $$  y < 1-\frac x3, $$
giving
$$ (x,y)\in B \iff 
\begin{cases} 
y\ge x\ge 0, y < 1-\frac x3,\\
 x\ge y\ge 0, x<1-\frac y3, \\ -y\ge x\ge 0, -y<1-\frac x3,\\
-x\ge y\ge 0, -x < 1-\frac y3,\\ 
\qquad \qquad \quad \vdots 
\end{cases}$$
Here's 6 of the octants (because Desmos has 6 colours) together with a full plot that relies on Desmos to interpret the implicit inequality:

