 Solution

For a complex derivative to exist, the difference quotient must agree along real and imaginary increments. These give $u_x+i{v}_x$ and $v_y-i{u}_y$, respectively. Equating their real and imaginary parts gives the Cauchy-Riemann equationsFor $g(z)=z\mathrm{Re}z$, 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,so $\boxed{g^{\prime }(0)=0}$. The function is differentiable at that isolated point but is not holomorphic on any neighborhood of it.