$\because$ $C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ for point ($\alpha$, $\alpha$) ${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$ $\therefore$ $\alpha = \sqrt 2$ On differentiating $({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ we get $x + yy' + 5{({x^2} - {y^2} - 1)^4}(x - yy') = 0$ ...... (i) When $x = y = \sqrt 2$ then $y' = {3 \over 2}$ Again on differentiating eq. (i) we get : $1 + {(y')^2} + yy'' + 20({x^2} - {y^2} - 1)(2x - 2yy')(x - y'y) + 5{({x^2} - {y^2} - 1)^4}(1 - y{'^2} - yy'') = 0$ For $x = y = \sqrt 2$ and $y' = {3 \over 2}$ we get $y'' = - {{23} \over {4\sqrt 2 }}$ $\therefore$ $3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$