It is evident that your functional $e(f, Q)$ has minimizer in the space $W^{1,2}$ since you have coercivity in this space and an application of direct method of calculus of variation givee you the required result in $W^{1,2}$. So the system of Euler-Lagrange equation in the above has solution in $W^{1,2}$. Now from the first equation (equation (i) you can compute $\dot{Q}$ and $Q$, in fact we have $$ \dot{Q} = \frac{1}{f^4}Q C$$ in which $C$ is a (skew symmetric) constant matrix. therefore $|\dot{Q}|^4 < \infty$ and hence $Q \in W^{1,4}$ (in fact by this argument you can show that the solution $Q$ is smooth). Similarly form the equation (ii) you can compute $f''$ in terms of $f$ and $f'$ and $|\dot{Q}|$ which shows that $f''$ exists (note that $f$ is bounded on [a,b]) so $f'$ is continuous and consequently for the solution $f\in W^{1,2}$ of the above system we have $f\in W^{1,4}$. In fact by the above argument one can show that the solution $f$ of the above system is also smooth (similar to the solution $Q$). Therefore the solution (Q,f) which first was proved to be in the space $W^{1,2}$ is in the space $W^{1,4}$ and moreover by applying the system of equation (i) and (ii) similarly we could prove that this solution $(Q, f)$is a $C^{\infty}$ solution.

It is evident that your functional $e(f, Q)$ has minimizer in the space $W^{1,2}$ since you have coercivity in this space and an application of direct method of calculus of variation givee you the required result in $W^{1,2}$. So the system of Euler-Lagrange equation you wrote in below, has solution in $W^{1,2}$. Now from the first equation (equation (i) you can compute $\dot{Q}$ and $Q$, in fact we have $$ \dot{Q} = \frac{1}{f^4}Q C$$ in which $C$ is a (skew symmetric) constant matrix. therefore $|\dot{Q}|^4 < \infty$ and hence $Q \in W^{1,4}$ (in fact by this argument you can show that the solution $Q$ is smooth). Similarly form the equation (ii) you can compute $f''$ in terms of $f$ and $f'$ and $|\dot{Q}|$ which shows that $f''$ exists (note that $f$ is bounded on [a,b]) so $f'$ is continuous and consequently for the solution $f\in W^{1,2}$ of the above system we have $f\in W^{1,4}$. In fact by the above argument one can show that the solution $f$ of the above system is also smooth (similar to the solution $Q$). Therefore the solution (Q,f) which first was proved to be in the space $W^{1,2}$ is in the space $W^{1,4}$ and moreover by applying the system of equation (i) and (ii) similarly we could prove that this solution $(Q, f)$is a $C^{\infty}$ solution.