= Solution
Take independent smooth <variations> $y+\varepsilon\eta$ and $z+\varepsilon\zeta$. Differentiating the <functional> and using <integration by parts> gives its <first variation>
$$
\delta J=[F_{y'}\eta+F_{z'}\zeta]_a^b+\int_a^b\left[(F_y-\tfrac d{dx}F_{y'})\eta+(F_z-\tfrac d{dx}F_{z'})\zeta\right]dx.
$$
First choose <variations> vanishing at the endpoints. To use the stated form of the <fundamental lemma of the calculus of variations>, take $\eta=(x-a)(b-x)\chi$ with arbitrary smooth $\chi$, and similarly for $\zeta$. The lemma makes $(x-a)(b-x)(F_y-dF_{y'}/dx)$ vanish, and hence makes the unweighted coefficient vanish in the interior; continuity extends this to the endpoints. Applying the same argument to $z$ gives the <Euler-Lagrange equations>
$$
\boxed{\frac d{dx}F_{y'}=F_y,\qquad\frac d{dx}F_{z'}=F_z.}
$$
If either endpoint value of $y$ or $z$ is free, the corresponding variation there is arbitrary. The remaining boundary term gives the <natural boundary conditions for a free endpoint>: $F_{y'}=0$ for a free $y$ value and $F_{z'}=0$ for a free $z$ value at that endpoint. With both coordinates free at both endpoints, all four conditions apply.
For the specified integrand these equations reduce to
$$
y''+5y-2z=0,\qquad z''+2y=0,
$$
and the natural conditions are $y'=z'=0$ at each free endpoint. Eliminating $z$ gives
$$
y^{(4)}+5y''+4y=0,\qquad (D^2+1)(D^2+4)y=0.
$$
Its four independent modes give the full general solution
$$
\begin{aligned}
y&=A\cos x+C\sin x+B\cos2x+D\sin2x,\\
z&=2A\cos x+2C\sin x+\tfrac12B\cos2x+\tfrac12D\sin2x.
\end{aligned}
$$
At zero, $y'(0)=C+2D$ and $z'(0)=2C+D$. Vanishing forces $C=D=0$, giving
$$
\boxed{y=A\cos x+B\cos2x,\qquad z=2A\cos x+\frac B2\cos2x.}
$$
At $x=\pi/6$ the two derivative conditions become $A/2+\sqrt3B=0$ and $A+\sqrt3B/2=0$. Their coefficient determinant is nonzero, so $A=B=0$. Thus \b[the unique stationary extremal with all four natural boundary conditions is $y=z=0$]. The word extremal here means a stationary point of the functional; the indefinite integrand does not require this point to be a minimum.
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