We have $g(x_1, x_2) = \frac{1}{6} \begin{pmatrix} x_1x_2+cx_2-1\\ x_1^2-x_2+1 \end{pmatrix}$
For what $c \in \mathbb{R}^+$ is the condition of the Banach fixed-point theorem in set $M = [-1,1]\times [-1,1]$ fulfilled?
My attempt was to do $||g(x_1,x_2)-g(y_1,y_2)||_\infty$ but I could not make it to the statement $k||x-y||_\infty$
Let us fix $x=(x_1,x_2)$, and $y=(y_1,y_2)$ in $M$. So we can use in estimations $|x_1|, |x_2|; |y_1|, |y_2|, \le 1$.
Let $a$ be the supremum norm of $x-y$, so $|x_1-y_1|\le a$ and $ |x_2-y_2|\le a$.
Then we have using brute force: $$ \begin{aligned} |(g(x)-g(y))_1| &=\frac 16|\ x_1x_2+cx_2-y_1y_2-cy_2\ | \\ &\le \frac 16|\ x_1x_2-y_1y_2\ | + \frac {|c|}6|x_2-y_2| \\ &= \frac 16|\ x_1x_2-x_1y_2+x_1y_2-y_1y_2\ | + \frac {|c|}6a \\ &\le \frac 16|x_1|\cdot |x_2-y_2| + \frac 16|y_2|\cdot |x_1-y_1| + \frac {|c|}6a \\ &= \left( \frac 16|x_1| + \frac 16|y_2| + \frac {|c|}6 \right)a \\ &\le \left( \frac 16+ \frac 16+ \frac {|c|}6 \right)a\ , \\[3mm] |(g(x)-g(y))_2| &=\frac 16|\ x_1^2-x_2-y_1^2+y_2\ | \\ &\le \frac 16|x_1+y_1|\cdot|x_1-y_1|+\frac 16|x_2-y_2| \\ &\le \left(\frac 16\cdot 2+\frac 16\right)a=\frac 12 a \ . \end{aligned} $$ From this, $$ \|\ g(x)-g(y)\ \|_\infty \le \max\left(\ \frac {2+|c|}6,\ \frac 12\ \right)\|x-y\|_\infty\ . $$ For $c\in(-4,4)$ we have thus a contraction. (The problem considers only $c>0$, but we have not used it.)
Do we need to find all $c$, so that we have a contraction?