Solution
= Solution
Each denominator is at least one and positive, so each component of $F$ lies in $(0,1]$. Taking the <maximum norm> therefore gives
$$
\boxed{\|F(x,Z)\|_\infty=\max\!\left\{\frac1{|z_2|+x^2+1},\frac1{|z_1|+1}\right\}\leq1.}
$$
This is a global bound, so in particular the constant $M=1$ is valid throughout the specified cube.