= Solution
For $a,b\geq0$ and $c\geq1$,
$$
\left|\frac1{a+c}-\frac1{b+c}\right|=\frac{|a-b|}{(a+c)(b+c)}\leq|a-b|.
$$
Apply this to the first component with $a=|z_2|$, $b=|w_2|$, $c=x^2+1$. The <reverse triangle inequality> gives
$$
|F_1(x,Z)-F_1(x,W)|\leq\big||z_2|-|w_2|\big|\leq|z_2-w_2|.
$$
For the second component use $a=|z_1|$, $b=|w_1|$, $c=1$ to obtain $|F_2(x,Z)-F_2(x,W)|\leq|z_1-w_1|$. Taking the maximum yields
$$
\boxed{\|F(x,Z)-F(x,W)\|_\infty\leq\|Z-W\|_\infty.}
$$
Thus $F$ is globally <Lipschitz continuous> in the dependent vector, uniformly in $x$, with <Lipschitz constant> $1$. Differentiability of the absolute-value function at zero is unnecessary.
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