= Solution
Write the admissible <variation> as $y_\varepsilon=y+\varepsilon\eta$, where $\eta(x)\to0$ at both ends. Differentiating the <functional> under the integral sign gives the <first variation>
$$
\delta I[y;\eta]
=\int_{-\infty}^{\infty}
\left(y'\eta'+U(y)U'(y)\eta\right)dx.
$$
After <integration by parts>, with the boundary term vanishing,
$$
\boxed{\delta I[y;\eta]
=\int_{-\infty}^{\infty}
\left[-y''+U(y)U'(y)\right]\eta\,dx}.
$$
Thus the <Euler-Lagrange equation> is
$$
y''=U(y)U'(y).
$$
A second differentiation gives the <second variation>
$$
\boxed{\delta^2I[y;\eta]
=\int_{-\infty}^{\infty}
\left\{(\eta')^2+\left[U'(y)^2+U(y)U''(y)\right]\eta^2\right\}dx}.
$$
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