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?

To show $d((x_1,y_1),(x_3,y_3)) \leq d((x_1,y_1),(x_2,y_2)) + d((x_2,y_2),(x_3,y_3))$
Note that since the metric is defined as being the max, it is greater than or equal to both options. That is:
$$
\begin{align}
|y_1-y_2|, |3x_1+y_1-3x_2-y_2| \,\leq& \ d((x_1,y_1),(x_2,y_2)) \\
|y_2-y_3|, |3x_2+y_2-3x_3-y_3| \,\leq& \ d((x_2,y_2),(x_3,y_3)) 
\end{align}
$$
$ \\ $
Case 1: $\ d((x_1,y_1),(x_3,y_3)) = |y_1-y_3| $
$|y_1-y_3| \leq |y_1-y_2| + |y_2 - y_3| \leq d((x_1,y_1),(x_2,y_2)) +d((x_2,y_2),(x_3,y_3)) $
Case 2 is the exact same.
