x½+y½=1,
½x-½+½y-½y'=0, ½y-½y'=-½x-½, y'=-x-½y½=-y½/x½,
-¼x-3/2-¼y-3/2y'+½y-½y"=0,
-x-3/2-y-3/2y'+2y-½y"=0,
2y-½y"=x-3/2+y-3/2y',
y"=½x-3/2y½+½y-1y',
y"=d2y/dx2=½x-3/2y½+½y-1(-y½/x½), so:
d2y/dx2=½y(y-½)/(x(x½))-½/(xy)½
d2y/dx2=½y/(x(xy)½)-½/(xy)½,
d2y/dx2=(1/(2(xy)½))((y/x)-1)=(y-x)/(2x(xy)½)) which resembles answer c (with a change of sign in the numerator).