= Solution
Use $\|w\|_{W^{1,\infty}}=\|w\|_\infty+\|w'\|_\infty$. Fix $u$. For $w$ small, all triples $(x,u+tw,u'+tw')$, $0\leq t\leq1$, lie in a fixed <compact set> up to null sets. The continuous derivatives $f_u,f_v$ are uniformly continuous there, with a modulus $\omega(s)\to0$ as $s\to0$. The integral form of the mean-value expansion gives a remainder bounded by
$$
(b-a)\omega(\|w\|_{W^{1,\infty}})(\|w\|_\infty+\|w'\|_\infty).
$$
The proposed linear derivative is bounded because its coefficient <functions> are bounded on that <compact set>. The remainder is little-oh of the <norm>, proving <Fréchet differentiability> and
$$
\boxed{DI(u)w=\int_a^b[f_v(x,u,u')w'+f_u(x,u,u')w]dx.}
$$
The derivative $u'$ only needs to exist almost everywhere; the uniform compact-range bound handles its measurable representative.
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