 Solution

Minimize the squared distance $F=x^2+y^2$, which has the same minimizers as the distance. The Lagrange multiplier equations for the constraint $G=x^2+xy-1=0$ areThe constraint gradient cannot vanish on this branch, since its second component is $x>0$. Eliminate $\lambda =2y/x$ to obtain $x^2=2xy+y^2$. Put $t=y/x$; then $t^2+2t-1=0$, so $t=-1\pm \sqrt{2}$. But the constraint says $x^2(1+t)=1$, requiring $1+t>0$ and selecting $t=\sqrt{2}-1$. It follows thatTo prove a global minimum on the whole $x>0$ branch, substitute $y=1/x-x$ into $F$:The arithmetic-geometric mean inequality proves this bound, with equality only when $2x^2=x^{-2}$, namely $x=2^{-1/4}$. Thus the stationary point is the unique closest point on the required branch. Numerically it is approximately $(0.840896,0.348311)$ and its distance is $0.910180$ in coordinate units.