= Solution
Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,
$$
u_x=\frac1{1+r}-\frac{x^2}{r(1+r)^2},\qquad v_y=\frac1{1+r}-\frac{y^2}{r(1+r)^2},\qquad u_y=v_x=-\frac{xy}{r(1+r)^2}.
$$
The two <Cauchy-Riemann equations> would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not <complex differentiable> at any nonzero point.
At zero, however,
$$
\frac{f(h)-f(0)}h=\frac1{1+|h|}\longrightarrow1.
$$
So \b[$f$ is <complex differentiable> at zero, with <derivative> one, but is not <holomorphic> on $\mathbb C$ or on any nonempty open subset]. <Complex differentiability> at a single point is weaker than <holomorphy> in a neighborhood.
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