Consider the functional $ F \left( u \right) = \int { \left( \frac{{d}^{2} u}{d{x}^{2}} \right) }^{2} dx $.
defining $ v = {u}_{x} $ the following is given:
$ F \left( u \right) = \int_{\Omega} {{u}_{xx}}^{2} dx = \int_{\Omega} {{v}_{x}}^{2} dx = G \left( v \right) $
Now, using the Gateaux derivative definition and $ L \left( x, v, {v}_{x} \right) = {{v}_{x}}^{2} $: $ {G}' \left( v \right) = \int_{\Omega} \left( \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) - \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) \right) h dx $
Hence the critical point happens at $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = 0 $ since $ \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) $ vanishes.
This implies the E-L is given by $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = \frac{d}{dx} 2 {v}_{x} = 2 {v}_{xx} = 0 $ which means $ {u}_{xxx} = 0 $.
This is probably a wrong answer. Yet it rose from a discussion with @Mark Peletier, hence I mark it as Community Wiki so people will be able to see why this property of the E-L can not be used here.
Thank You.