= Solution
Holding $y$ fixed, the <chain rule> gives
$$
\boxed{\left(\frac{\partial v}{\partial u}\right)_y\left(\frac{\partial u}{\partial x}\right)_y=\left(\frac{\partial v}{\partial x}\right)_y,}
$$
where the local change of variable from $x$ to $u$ is valid.
For the <internal energy> $U(S,V)$, its <exact differential> identifies $T=U_S$ and $p=-U_V$. The <Maxwell relation> from equality of mixed <partial derivatives> is $T_V|_S=-p_S|_V$. At fixed <volume (thermodynamics)>, the <constant-volume heat capacity> satisfies
$$
C_V=\left(\frac{\partial U}{\partial T}\right)_V=T\left(\frac{\partial S}{\partial T}\right)_V,
$$
so $T_S|_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the <chain rule> to $p(T,V)$ gives the <isentropic temperature-volume derivative>
$$
\boxed{\left(\frac{\partial T}{\partial V}\right)_S=-\left(\frac{\partial p}{\partial T}\right)_V\left(\frac{\partial T}{\partial S}\right)_V=-\frac{T}{C_V}\left(\frac{\partial p}{\partial T}\right)_V.}
$$
For the stated one-mole <ideal gas>, the <equation of state> gives $p_T|_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an <isentropic process>, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:
$$
\boxed{\left(\frac{\partial T}{\partial V}\right)_S=-\frac{2T}{3V}.}
$$
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