if $f(x,y,y')$ does not depend on $x$ then the Euler-Lagrange equation becomes in:
$$f(x,u,u')-u'f_{y'}(x,u,u')=c $$ in $[x_0,x_1]$ for some $c \in \mathbb{R}$
Ok I need prove this, I know that if $u$ is a local minimum then
$$f_{y}(x,u,u')-\frac{d}{dx}(f_{y}(x,u,u'))=0 $$ in $[x_0,x_1]$
Given $f(x,y,y')$ we have
$$ \frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ (1) $$
and also
$$ \frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'\right)=\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ \ (2) $$
and according the Euler-Lagrange equation we have
$$ \frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)=0\ \ \ \ \ \ \ \ \ \ \ (3) $$
then using $(3)$ into $(2)$ we obtain
$$ \frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'\right)=\frac{\partial f}{\partial y}y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ \ (4) $$
now from $(1),(4)$
$$ \frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'-f\right)=-\frac{\partial f}{\partial x} $$
but $\frac{\partial f}{\partial x} = 0$ because $f=f(y,y')$ hence
$$ \frac{\partial f}{\partial y'}y'-f = C $$