Using Euler-Lagrange equations Let $$F[y,z]=\int_a^b \left[y'(t)z(t)-\frac{1}{2}y(t)^2-\frac{1}{2}z(t)^2 \, \right]dt$$ and let $$L[x]=\int_a^b f(x(t),x'(t)) \, dt$$
I'm told that if $\tilde{x}$ is a stationary point of $L$ then $(\tilde{x},\tilde{x}')$ is a stationary point of $F$. I'm now asked to find $f$.
I have considered the Euler-Lagrange equations for $F$ and found that $y'=z$ and $z'=-y$. Using the Beltrami equation I have also found that $\frac{1}{2}y^2+\frac{1}{2}z^2=c \text{ (constant)}$. Putting this all together gives me the differential equation $(y')^2+y^2=c$. 
I'm not sure if this helps me in any way but I don't know what else I can do here.
 A: It is the Lagrangian for an unidimensional harmonic oscillator of a particle of unit mass. $$f(x,x')=\frac{1}{2}(x')^2-\frac{1}{2}x^2$$
The ODE you've found is the conservation of mechanical energy for this system.
I've found the answer with a simple substitution at F.
Added
$$F[x,x']=\int_a^b x'(t)x'(t)-\frac{1}{2}x^2-\frac{1}{2}(x')^2 \, dt=\int_a^b \frac{1}{2}(x')^2-\frac{1}{2}(x^2) \, dt$$
Now, if being stationary for $L$ must imply being stationary for $F$, the integrands must differ in the total derivative with respect to time of any function of coordinate and time. I chose this to be zero, but it can be done well:
$$f(x,x')=\frac{1}{2}(x')^2-\frac{1}{2}x^2+\frac{\mathbb d}{\mathbb dt}g(x,t)$$
A: Hint: One idea is to "integrate out the $z$-variable", i.e. to first find the stationary point wrt. the $z$ variable. Completing the square wrt. the $z$ variable, OP's functional becomes
$$  F[y,z]~=~\int_a^b \! \mathrm{d}t\left(y^{\prime} z -\frac{1}{2}y^2-\frac{1}{2}z^2 \right)~=~\frac{1}{2}\int_a^b \! \mathrm{d}t\left(y^{\prime 2}  -y^2 \right) -  \frac{1}{2}\int_a^b \! \mathrm{d}t\left(z-y^{\prime}   \right)^2. \tag{1}$$
Hence
$$ \max_{z}F[y,z]~=~ \frac{1}{2}\int_a^b \! \mathrm{d}t\left(y^{\prime 2}  -y^2 \right) .\tag{2} $$
From eq. (2) we can directly read off a possible $f$-function:
$$f(x,x^{\prime})~=~\frac{1}{2}\left(x^{\prime 2}  -x^2 \right) .\tag{3} $$
Note that $f$ is far from unique: For starters, we may scale $f$ with an overall constant or add a total derivative term. But more exotic modifications are also possible, cf. e.g. this Math.SE post.
