= Solution
At fixed particle number, the differential of the <internal energy> identifies $T=(\partial U/\partial S)_V$ and $p=-(\partial U/\partial V)_S$. Equality of continuous <mixed partial derivatives> gives the first <Maxwell relation>:
$$
\boxed{\left(\frac{\partial T}{\partial V}\right)_S=\frac{\partial^2U}{\partial V\partial S}=-\left(\frac{\partial p}{\partial S}\right)_V.}
$$
For the <enthalpy>, $dH=dU+p\,dV+V\,dp=T\,dS+V\,dp$. Thus $T=(\partial H/\partial S)_p$, $V=(\partial H/\partial p)_S$, and the same argument gives the <enthalpy Maxwell relation>
$$
\boxed{\left(\frac{\partial T}{\partial p}\right)_S=\frac{\partial^2H}{\partial p\partial S}=\left(\frac{\partial V}{\partial S}\right)_p.}
$$
The smoothness and independent thermodynamic coordinates are understood within a single phase.
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