$$\because$$ $$C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$$ 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 $$y'' = - {{23} \over {4\sqrt 2 }}$$

$$\therefore$$ $$3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$$