= Solution
For a fixed amount of a <simple compressible system>, write the <internal energy> as $U(S,V)$. Its exact <differential> is
$$
dU=T\,dS-p\,dV,\qquad T=\left(\frac{\partial U}{\partial S}\right)_V,\qquad -p=\left(\frac{\partial U}{\partial V}\right)_S.
$$
Assuming the <thermodynamic potential> has continuous second <partial derivatives>, equality of its <mixed partial derivatives> yields 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 second <Maxwell relation>, use the <Legendre transform> $H=U+pV$, the <enthalpy>. Its natural variables are $S,p$ since
$$
dH=dU+p\,dV+V\,dp=T\,dS+V\,dp.
$$
Thus $T=(\partial H/\partial S)_p$ and $V=(\partial H/\partial p)_S$. Equality of the <mixed partial derivatives> now gives
$$
\boxed{\left(\frac{\partial T}{\partial p}\right)_S=\frac{\partial^2H}{\partial p\partial S}=\left(\frac{\partial V}{\partial S}\right)_p.}
$$
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