= Solution
Differentiating $f(\mu,\sigma)=\mathbb E[U(\mu+\sigma Z)]$ twice gives the <Hessian matrix>
$$
\operatorname{Hess}f
=\begin{pmatrix}
\mathbb E[U''(X)]&\mathbb E[ZU''(X)]\\
\mathbb E[ZU''(X)]&\mathbb E[Z^2U''(X)]
\end{pmatrix}
=\mathbb E\left[U''(X)
\begin{pmatrix}1\\Z\end{pmatrix}
\begin{pmatrix}1&Z\end{pmatrix}\right].
$$
For every vector $(a,b)$,
$$
\begin{pmatrix}a&b\end{pmatrix}
(\operatorname{Hess}f)
\begin{pmatrix}a\\b\end{pmatrix}
=\mathbb E[U''(X)(a+bZ)^2]\leq0.
$$
Thus the Hessian is a <negative semidefinite matrix>, and
$$
\boxed{f\text{ is jointly concave in }(\mu,\sigma)}.
$$
Equivalently, with $W=-U''(X)\geq0$, its determinant is
$$
\mathbb EW\,\mathbb E[Z^2W]-(\mathbb E[ZW])^2\geq0
$$
by the <Cauchy-Schwarz inequality>, while both diagonal entries are nonpositive.
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