Proving that $d$ defines a metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow [0,\infty)$ I have to show that $d((x_1,y_1),(x_2,y_2)) = max\{|3x_1 + y_1 − 3x_2 − y_2|, |y_1 − y_2|\}$ defines a metric.
My approach:
Let $x=(x_1,y_1)$ and $y = (x_2,y_2)$
$M_1: d(x,y) \geq 0$ is clearly satisfied since $d$ chooses the maximum value between two values defined by an absolute value.
$$d(x,y) = 0 \iff  max\{|3x_1 + y_1 − 3x_2 − y_2|, |y_1 − y_2|\} = 0 \\
\iff |3x_1 + y_1 − 3x_2 − y_2|= 0 \land |y_1 − y_2| = 0 \\
\iff y_1 = y_2 \land x_1 = x_2.$$
$M_2: d(x,y) = |3x_1 + y_1 − 3x_2 − y_2|$ or $d(x,y) = |y_1 − y_2|$ and in both cases we can add a minus inside the absolute value and get that $d(x,y)=d(y,x)$.
Are these correct? 
I am not sure how to go about $M_3$ since I think there are a lot of cases. Any help?

Introduce new coordinates $(u,v)$ in the $(x,y)$-space $V:={\mathbb R}^2$, related to $x$, $y$ by
$$u:=3x+y,\quad v:=y,\qquad{\rm resp.,}\qquad x={1\over3}(u-v),\quad y=v\ .$$
In terms of the new cordinates the metric in $V$ is just the metric induced by the $\ell_\infty$-norm
$$\|(u,v)\|:=\max\bigl\{|u|,|v|\bigr\}\ .$$
