For the curve $C:\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0$, the value of $3 y^{\prime}-y^{3} y^{\prime \prime}$, at the point $(\alpha, \alpha)$, $\alpha>0$, on C, is equal to ____________.

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