= Solution
Under a <gauge transformation>, the <electromagnetic field tensor> changes by
$$
\delta F_{\mu\nu}=\partial_\mu\partial_\nu\xi-\partial_\nu\partial_\mu\xi=0.
$$
The partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the <Maxwell Lagrangian> depends only on a gauge-invariant tensor:
$$
\boxed{F'_{\mu\nu}=F_{\mu\nu},\qquad\mathcal L'=-\frac14F'_{\mu\nu}F'^{\mu\nu}=\mathcal L.}
$$
Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.
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