= Solution
For every $z\in B$, $z_3+3>2$ and $|z_1+z_2|<2$. Consequently each of the three <partial derivatives> in the preceding <gradient> has absolute value below $1/2$, so
$$
\|\nabla g(z)\|^2<\frac14+\frac14+\frac14=\frac34.
$$
The segment joining $x,y\in B$ lies inside $B$. The <mean value theorem> and <Cauchy-Schwarz inequality> therefore give
$$
|g(y)-g(x)|\leq\|\nabla g(x+s(y-x))\|\,\|y-x\|\leq\boxed{\frac{\sqrt3}{2}\|y-x\|}.
$$
For $x=y$ both sides vanish. This proves the required <Lipschitz continuity> uniformly on the ball.
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