Given that ‌x√1+y+y√1+x=0 . . . (i) ‌ or ‌‌‌x√1+y=−y√1+x On squaring both sides, we get ‌x2(1+y)=y2(1+x) ⇒x2−y2+x2y−xy2=0 ⇒(x−y)(x+y)+xy(x−y)=0 ⇒(x−y)(x+y+xy)=0 x−y≠0 because it does not satisfy the Eq. (i) ‌∴‌‌x+y+xy=0 ‌⇒‌‌y=−‌