= Solution
For fixed endpoint values, the <Euler-Lagrange equation> is $f_y-df_{y'}/dx=0$. Along a stationary curve, the <chain rule> gives
$$
\frac{d}{dx}(y'f_{y'}-f)=y'\left(\frac{df_{y'}}{dx}-f_y\right)-f_x.
$$
When $f_x=0$, the <Beltrami identity> therefore takes the requested sign convention
$$
\boxed{y'f_{y'}-f=A.}
$$
By <Fermat principle>, the elapsed-time integrand for the given speed is $f=c_0^{-1}\sqrt{1-ky}\sqrt{1+(y')^2}$. Thus
$$
y'f_{y'}-f=-\frac{\sqrt{1-ky}}{c_0\sqrt{1+(y')^2}}=A.
$$
The constant is negative for a finite regular ray. Squaring gives $(1-ky)/(1+(y')^2)=c_0^2A^2$. At a maximum height $y_0$, the slope is zero, so $c_0^2A^2=1-ky_0>0$. Hence
$$
(y')^2=\frac{k(y_0-y)}{1-ky_0}.
$$
Separate variables on either side of the turning point. Integration gives the <parabolic light rays in a reciprocal-square-root speed profile>
$$
\boxed{y(x)=y_0-\frac{k(x-x_c)^2}{4(1-ky_0)}.}
$$
The quadratic opens downwards, with its greatest height $y_0$ at $x=x_c$. The positive index condition holds throughout the layer because $kh<1$. Vertical rays are a separate zero-horizontal-momentum case, outside the regular graph representation $y(x)$; they are not returning rays of the type considered below.
Equal endpoint heights at $x=\pm x_0$ force $x_c=0$. Setting $y(\pm x_0)=0$ gives
$$
\boxed{(kx_0)^2=4ky_0(1-ky_0).}
$$
For nonzero separation, a real returning ray requires $kx_0\leq1$ and has candidate turning heights
$$
\boxed{y_0=\frac{1\pm\sqrt{1-k^2x_0^2}}{2k}.}
$$
Only choices satisfying $0<y_0<h$ are rays wholly within the specified layer. Depending on its thickness there may be two, one, or no admissible choices; at zero discriminant the two algebraic candidates coincide. The entry and exit points at $y=0$ are interpreted as limiting boundary points of the open medium.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1-q9-parabolic-rays.png]
{title=Two possible returning parabolic rays, subject to the optical layer height}
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