I'm interested in minimizing the functional $I[f]=\int_{a}^{b} \mathcal{L}(x,f,f',f'',f^{-1},(f^{-1})', (f^{-1})'') dx$.
Would it be allowable to consider that $f^{-1}$ is some function $g$ and apply the Euler-Lagrange Equation with several functions and higher derivatives? I.e., are the following derivations correct (necessary, but need not be sufficient), considering that $g=f^{-1}$? If not, what would be the right variational equations to capture this problem "easily"?
$$ \cfrac{\partial \mathcal{L}}{\partial f} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial f'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial f''}\right) = 0 $$
and
$$ \cfrac{\partial \mathcal{L}}{\partial g} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial g'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial g''}\right) = 0 $$
Note: When I'm denoting the partials, e.g. $\cfrac{\partial \mathcal{L}}{\partial f}$, I literally mean only with respect to the corresponding symbol, as in the original Euler-Lagrange notation. If (for example only) I had that $\mathcal{L}=(f(x))^2+(f^{-1}(x))^3 + f'(x) + 2(f^{-1})'(x)$, then it would be that $\cfrac{\partial \mathcal{L}}{\partial f} = 2f(x)$, $\cfrac{\partial \mathcal{L}}{\partial f'} = 1$, $\cfrac{\partial \mathcal{L}}{\partial g} = 3(f^{-1}(x))^2$, and $\cfrac{\partial \mathcal{L}}{\partial g'} = 2$.
Note 2: In fact, I'm interested in finding a function such that $f=f^{-1}$ and the above functional form $\mathcal{L}$ has arisen from incorporating a Lagrange multiplier: $+\lambda (f(x) - f^{-1}(x))$ and some other additional factors relevant to that (so that we don't get trivialities like $\lambda=0$ at the saddle point), i.e., originally I had $\mathcal{L}(x,f,f',f'')$ but with the additional constraint that $f=f^{-1}$. If this more specific problem can be captured in an easier way than the above, more general one, then it would be preferable.