I'm supposed to show, that if my function $f(y,y',x)$ is independent of $y$, then the Euler-Lagrange equation turns out to be $\partial f/\partial y' = const$.
Now, the Euler-Lagrange equation is given by:
$$\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0$$
So, it's easy to see, that if it's independent of $y$, then the first term just becomes zero, and you are left with $$\frac{d}{dx}\frac{\partial f}{\partial y'} = 0$$.
But is that the proof ? When the first terms is equal to zero, then $\frac{\partial f}{\partial y'} = const$ ?
Don't I have to take the $\frac{d}{dx}$ into account, or...?
Your statement is true. If $f$ is independent of $y$ then the Euler-Lagrange equation becomes ${{d}\over{dx}}\left( {{\partial f}\over{\partial y'} }\right)=0$. So you have that the derivative of function ($ {{\partial f}\over{\partial y'}}$) is zero. Which function has its derivative zero? The answer would be a constant function. As a result, ${{\partial f}\over{\partial y'}}=const$.