= Solution
Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y'^2}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y'$, use the full <Euler-Lagrange equation>:
$$
yy''+y'^2+1=0,\qquad (yy')'=-1.
$$
Two integrations give $y^2=-(x-x_0)^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. Thus
$$
\boxed{y(x)=\sqrt{2-x^2},\qquad -1\leq x\leq1.}
$$
The path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $yy''+y'^2+1=0$, so is not another extremum.
One can prove global minimality, rather than only stationarity. Define $\Phi(x,y)=\operatorname{arsinh}(x/y)$ on the upper half-plane. Then
$$
\nabla\Phi=\frac{(1,-x/y)}{\sqrt{x^2+y^2}},\qquad |\nabla\Phi|=\frac1y.
$$
The <Cauchy-Schwarz inequality> gives, along any admissible path from $A$ to $B$,
$$
D=\int\frac{ds}{y}\geq\int\nabla\Phi\cdot d\mathbf r
=\Phi(1,1)-\Phi(-1,1)=2\operatorname{arsinh}1.
$$
Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This <calibration for semicircular hyperbolic geodesics> proves
$$
\boxed{D_{\min}=2\operatorname{arsinh}1=2\ln(1+\sqrt2).}
$$
The geometry is that of a <hyperbolic geodesic> in the <Poincaré half-plane model>. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.
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