= Solution
An <immersion> $\phi:U\to\mathbb R^3$ has differential of rank $2$, equivalently <linearly independent> $\phi_u,\phi_v$. It has <isothermal coordinates> when its <first fundamental form> has $E=G>0$, $F=0$. A <minimal surface> has zero <mean curvature>.
For an isothermal <immersion>, differentiation of $E=G$ and $F=0$ gives
$$
(\phi_{uu}+\phi_{vv})\cdot\phi_u=\tfrac12E_u+F_v-\tfrac12G_u=0,
$$
and similarly its scalar product with $\phi_v$ is zero. Its normal component is $e+g$. The given <mean curvature> formula reduces to $H=(e+g)/(2E)$, so
$$
\boxed{\phi_{uu}+\phi_{vv}=2EH\mathbf n.}
$$
Thus an isothermal <immersion> is minimal exactly when each coordinate is a <harmonic function>.
An explicit example is the <Enneper surface>, parametrized on $\mathbb R^2$ by
$$
\phi(u,v)=(u-u^3/3+uv^2,\ v-v^3/3+u^2v,\ u^2-v^2).
$$
Every coordinate has zero Laplacian. Direct differentiation gives $\phi_u\cdot\phi_v=0$ and $|\phi_u|^2=|\phi_v|^2=(1+u^2+v^2)^2>0$. Thus it is an isothermal <immersion> and is minimal everywhere.
It is not an open subset of any of the three listed surfaces. The parameter points $(\sqrt3,0)$ and $(-\sqrt3,0)$ both map to $(0,0,3)$, but their tangent planes differ: $\phi_u=(-2,0,\pm2\sqrt3)$ while $\phi_v=(0,4,0)$. A plane, <catenoid>, or <helicoid> is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank-$2$ tangent planes at this point would have to equal its tangent plane, a contradiction.
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