Suppose that $f(x,y(x),y'(x))$ is s.t $f\in{C^2}$ and $y'(x)\ne{0}.$
I am trying to show that the Euler-Lagrange equation $\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0$ is equivalent to $\frac{\partial f}{\partial x}-\frac{d}{dx}\big(f-y'\frac{\partial f}{\partial y'}\big) = 0$.
Clearly $f\in{C^2}$ means all the second partial derivatives of $f$ exist and $y'(x)\ne{0}$ is probably necessary for some division by $y'$. I don't know if this is useful but I tried considering when $f-y'\frac{\partial f}{\partial y'}=\frac{\partial f}{\partial y'}.$
Can you give me a possible hint to start me off.
If we take the total derivative of $f$ w.r.t. $x$ we get that $$\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial f}{\partial y'}\frac{\mathrm{d}y'}{\mathrm{d}x}$$