Is it a contraction map? I have that map 
$$
f:(\mathbb{R}^2,d_1)\to(\mathbb{R}^2,d_1)\\
(x,y)\mapsto \left(y-\frac13 \tanh(x)+\frac14 Argsh(y),\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 4x-\tanh(y)+\frac43 Argsh(x)\right)
$$
where $d_1((x,y),(x',y'))=|x-x'|+|y+y'|$
I calcultate $$ d_1(f(x,y),f(x',y'))\leq \frac{43}{12}|y-y'|+\frac{13}{3}|x-x'|$$ like this it is not contraction 
Can someone tel me if what i do is correct ?
$$
d_1(f(x,y),f(x',y'))\leq |y-y'|+\frac13|\tanh(x)-\tanh(x')|+\frac14 |Argsh(y)-Argsh(y')|+4|x-x'|+|\tanh(y')-\tanh(y)|+\frac43 |Argsh(x)-Argsh(x')|\\ \leq |y-y'|+\frac13|x-x'|+\frac14|y-y'|+4|x-x'|+|y'-y|+\frac43|x-x'|$$
Thank you 

You work with the metric $d_1((x,y),(x',y') = \lvert x - x' \rvert + \lvert y - y' \rvert $ (typo in your question!). Your inequality
$$d_1(f(x,y),f(x',y')) \le 43/12\lvert y - y' \rvert +  13/3\lvert x - x' \rvert$$
does not help because you cannot be sure that it best possible.
It suffices to look at special values. Let $(x,y) = (a,0)$ with $a >0$ and $(x',y') = (0,0)$. We have $f(a,0) = (-1/3\tanh(a),4a + 4/3Argsh(a))$ and $f(0,0) = (0,0)$. The distance between these two points is $1/3 \tanh(a) + 4a + 4/3 Argsh(a) > 4a$. Therefore $f$ is no contraction.
One more remark: $Argsh$ seems to be used only in French literature. It is the inverse of the hyperbolic sine $\sinh$ and is most frequently denoted by $arsinh$.
