Let $F=F\left(y, y^\prime, y^{\prime\prime}, x\right) = F\left(y^\prime, y^{\prime\prime}\right)$ and define $$H = H\left(y^\prime, y^{\prime\prime}\right) = y^{\prime\prime}\frac{\partial F}{\partial y^{\prime\prime}} - y^\prime \left(\frac{d}{dx}\frac{\partial F}{\partial y^{\prime\prime}}-\frac{\partial F}{\partial y^\prime}\right) - F$$ Show that at the stationary solution we have $H = \text{constant}$.
Hint: compute $dF/dx$ and $dH/dx$ and use the fact that $$\frac{d}{dx}\frac{\partial F}{\partial y^{\prime\prime}} - \frac{\partial F}{\partial y^\prime} = \text{constant}$$
So far I have $$\frac{dF}{dx}=\frac{d}{dx}\frac{\partial F}{\partial y^\prime} + \frac{d}{dx}\frac{\partial F}{\partial y^{\prime\prime}}$$ $$\frac{dH}{dx}=\frac{d}{dx}\frac{\partial H}{\partial y^\prime} + \frac{d}{dx}\frac{\partial H}{\partial y^{\prime\prime}}$$ and $$ H = y^{\prime\prime}\frac{\partial F}{\partial y^{\prime\prime}} - ky^\prime - F $$ where $k$ is a constant.
By the Euler-Lagrange equation I know that the stationary solution must satisfy $$\frac{d}{dx}\frac{\partial H}{\partial y^\prime}-\frac{d^2}{dx^2}\frac{\partial H}{\partial y^{\prime\prime}}=0$$
However, I don't see yet how to combine all of this information. Hint please!
Hint: Your equations for $dF/dx$ and $dH/dx$ are not correct. Instead
$$ \frac{dF}{dx}=\frac{\partial F}{\partial y'} y'' + \frac{\partial F}{\partial y''} y''' \ \ \mbox{and} \ \ \ \frac{dH}{dx}=\frac{\partial H}{\partial y'} y'' + \frac{\partial H}{\partial y''} y''' $$ Using this, things cancel out nicely.