= Solution
An <isothermal parametrization> is an <immersion> with <first fundamental form> $E\,du^2+2F\,du\,dv+G\,dv^2=\lambda^2(du^2+dv^2)$, where $\lambda>0$. Thus $E=G=\lambda^2$ and $F=0$.
Write $W=\phi_{uu}+\phi_{vv}$. Differentiating the metric identities gives
$$
W\cdot\phi_u=\tfrac12E_u-\tfrac12G_u=0,\qquad
W\cdot\phi_v=-\tfrac12E_v+\tfrac12G_v=0.
$$
For example $\phi_{vv}\cdot\phi_u=(\phi_v\cdot\phi_u)_v-\phi_v\cdot\phi_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied <mean curvature> formula reduces to $H=(e+g)/(2\lambda^2)$. Therefore
$$
\boxed{\phi_{uu}+\phi_{vv}=2\lambda^2HN=2\lambda^2\mathbf H}.
$$
A surface is a <minimal surface> when its <mean curvature> <vector> vanishes. Since $\lambda^2>0$, the preceding identity proves \b[minimality is equivalent to $\Delta\phi=0$ in <isothermal coordinates>].
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