= Solution
Along a sufficiently smooth stationary curve, differentiation and the <Euler-Lagrange equation> give the <Beltrami identity>:
$$
\frac d{dx}(f-y'f_{y'})=f_yy'-y'\frac d{dx}f_{y'}=0,
\qquad\boxed{f-y'f_{y'}=\text{constant}.}
$$
The cancellation uses the absence of explicit $x$ dependence.
For an axisymmetric <surface of revolution>, its area is $I[y]=2\pi\int_{-l}^l y\sqrt{1+y'^2}\,dx$. A smooth positive-radius minimizer is stationary, so the <Beltrami identity>, with the constant factor omitted, gives
$$
\frac{y}{\sqrt{1+y'^2}}=k>0.
$$
Equivalently the <Euler-Lagrange equation> gives $yy''=1+y'^2$, whence $y''=y/k^2$. The first integral selects
$$
y=k\cosh((x-x_*)/k).
$$
Equal radii at $\pm l$ imply $x_*=0$ (subtract the two cosh values), giving
$$
\boxed{y(x)=k\cosh(x/k),\qquad r=k\cosh(l/k).}
$$
This is a <catenoid>. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.
Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow0$ and $a\to\infty$. There are exactly two positive values of $k$ when
$$
r/l>\min_{a>0}\frac{\cosh a}{a}\simeq1.50888.
$$
At equality there is one value, and below it there is none.
To match the printed <second variation> convention, write $I[y+\varepsilon\eta]=I[y]+\varepsilon\delta I+\varepsilon^2\delta^2I+o(\varepsilon^2)$; its $\delta^2I$ is the quadratic coefficient, half the second derivative with respect to $\varepsilon$. Expanding the area integrand gives
$$
\delta^2I=\pi\int_{-l}^l\left[\frac{y}{(1+y'^2)^{3/2}}\eta'^2
+2\frac{y'}{\sqrt{1+y'^2}}\eta\eta'\right]dx.
$$
On the <catenoid> these coefficients are $k\operatorname{sech}^2(x/k)$ and $2\tanh(x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta(\pm l)=0$. Since $(\tanh(x/k))'=\operatorname{sech}^2(x/k)/k$, this yields
$$
\boxed{\delta^2I=\pi\int_{-l}^l\left(k\eta'^2-\frac{\eta^2}{k}\right)\operatorname{sech}^2(x/k)\,dx.}
$$
This is the <axisymmetric second variation of a catenoid>, with the explicitly stated quadratic-coefficient convention.
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