It follows from
$\begin{align*}dU = -PdV + TdS + \mu dN\end{align*}$
that
$\begin{align*}dS = -\frac{PdV}{T} + \frac{dU}{T} - \frac{\mu dN}{T}\end{align*}$
So,
$\begin{align*}\left(\frac{\partial S}{\partial U} \right)_{V,N} = \frac{1}{T}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1986 Problem 66