The example I'm doing gives an equation $L(y, y') = \frac{y'}{y}$
then $\frac{\partial L}{\partial y} = -\frac{y'}{y^2}$ and $\frac{\partial L}{\partial y'} = \frac{1}{y}$ ... $\frac{d}{dx}\frac{\partial L}{\partial y'} = \frac{d}{dx}\frac{1}{y} = -\frac{y'}{y^2}$
substituting into the Euler Lagrange $\frac{\partial L}{\partial y} = \frac{d}{dx} \frac{\partial L}{\partial y'}$ then yields $-\frac{y'}{y^2} = -\frac{y'}{y^2}$
Does this mean that any differentiable function $y$ is a solution to the Euler-Lagrange equation for that Lagrangian??