= Solution
Vary the <action> $S=\int dt\,dx\,\mathcal L$ with compactly supported $\delta\phi$, or impose vanishing $\delta\phi$ and its first derivatives on the spacetime boundary. The <first variation> is
$$
\delta S=\int dt\,dx\left[\mathcal L_\phi\delta\phi+\mathcal L_{\phi_{,\mu}}\partial_\mu\delta\phi+\mathcal L_{\phi_{,\mu\nu}}\partial_\mu\partial_\nu\delta\phi\right].
$$
One <integration by parts> on the first-derivative term and two on the second-derivative term remove all <derivatives> from $\delta\phi$. The boundary terms vanish, leaving
$$
\delta S=\int dt\,dx\left[\mathcal L_\phi-\partial_\mu\mathcal L_{\phi_{,\mu}}+\partial_\mu\partial_\nu\mathcal L_{\phi_{,\mu\nu}}\right]\delta\phi.
$$
The <fundamental lemma of the calculus of variations> gives the <Euler-Lagrange field equation with second derivatives>:
$$
\boxed{\frac{\partial\mathcal L}{\partial\phi}-\partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}+\partial_\mu\partial_\nu\frac{\partial\mathcal L}{\partial(\partial_\mu\partial_\nu\phi)}=0.}
$$
Repeated indices run over time and space. The coefficient of the symmetric second derivative is defined by the displayed variation; if $\phi_{tx}$ is instead treated as a single independent mixed derivative variable, it is counted once rather than twice.
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