= Solution
For a complex derivative to exist, the difference quotient must agree along real and imaginary increments. These give $u_x+iv_x$ and $v_y-iu_y$, respectively. Equating their real and imaginary parts gives the <Cauchy-Riemann equations>
$$
\boxed{u_x=v_y,\qquad u_y=-v_x.}
$$
For $g(z)=z\operatorname{Re}z$, the components are $u=x^2$ and $v=xy$. The first equation requires $2x=x$, so $x=0$; the second requires $0=-y$, so $y=0$. Thus zero is the only possible differentiability point. Directly,
$$
\frac{g(h)-g(0)}h=\operatorname{Re}h\longrightarrow0,
$$
so $\boxed{g'(0)=0}$. The function is differentiable at that isolated point but is not <holomorphic> on any neighborhood of it.
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