I have the following lemma :
Let $f\in \mathcal C^1([a,b]\times \mathbb R\times \mathbb R)$, $f=f(x,u,\xi)$ satisfy some hypothesis (but they are not important for my question). Then any solution $u\in W^{1,p}(a,b)$ of $$\int_a^b [f_u(x,u(x),u'(x))v(x)+f_\xi(x,u(x),u'(x))v'(x)dx,\quad \forall v\in \mathcal C_0^\infty (a,b),$$ belong to $W^{1,\infty }(a,b)$ and the equation hold a.e., i.e. $$\frac{d}{dx}f_\xi=f_u,\quad a.e.$$
The proof goes like : Let $\varphi(x)=f_\xi(x,u(x),u'(x))$ and $\psi(x)=f_u(x,u(x),u'(x))$. We easily see that $\varphi\in W^{1,1}(a,b)$ and $\varphi'(x)=\psi(x)$ fo almost every $x\in (a,b)$, which mean that $$\frac{d}{dx}f_\xi=f_u,\quad a.e.$$ hold.
Question To be honnest, I don't really understand the argument. Indeed, to me, by an integration by part, we have that $$\int_a^b\left(f_u-\frac{d}{dx}f_\xi\right)v=0$$
for all $v\in \mathcal C_0^\infty (\Omega )$ and thus the claim follow by the fundamental lemma of the calculus of variation. So why doing all this ?
Your argument would have been correct if $f$ would have been $\mathcal C^2$. Unfortunately it's not the case. That's why you need that $\varphi\in W^{1,1}$ ; what tells you that $\varphi$ is weakly derivable, and thus allow you to integrate by part $$\int_a^b \varphi v'=-\int_a^b v\psi.$$