√(x/a)+√(y/b)=1, so differentiating:
½(1/√(ax)+y'/√(by)=0, y'=-√(by/ax), y'2=by/(ax) and 1+y'2=(ax+by)/(ax);
Radius of curvature r=(1+(dy/dx)2)3/2/d2y/dx2=(1+y'2)3/2/y'', using y' for dy/dx and y'' for d2y/dx2.
1/√(ax)+y'/√(by)=0 needs to be differentiated again to find the second derivative.
-½(1/√(ax3))-½(y'2/√(by3))+y''/√(by)=0,
y''/√(by)=½(1/√(ax3)+y'2/√(by3))=½(1/√(ax3)+(by/(ax))/√(by3)) (substituting y'2=by/(ax)),
y''=(√(by)/2)(1/√(ax3)+(√b/(ax))/√y)=½(√(by/ax3)+b/(ax))=(1/(2√(ax)))(√(by)/x+b/√(ax)),
y''=(1/(2√(ax)))(√(abxy)+bx)/(x√(ax)), 1/y''=2ax2/(√(abxy)+bx).
r=[(ax+by)/(ax)]3/2[2ax2/(√(abxy)+bx)]=2(ax+by)3/2[√(x/a)/(√(abxy)+bx)],
r=2(ax+by)3/2[√x/√a)/(√x(√(aby)+b√x)]=2(ax+by)3/2/(ab√(y/b)+ab√(x/a))=2(ax+by)3/2/(ab(√(y/b)+√(x/a)).
Since √(x/a)+√(y/b)=1, r=2(ax+by)3/2/(ab). QED