For the curve C: ( x ^ 2 + y ^ 2 - 3) + ( x ^ 2 - y ^ 2 - 1) ^ 5 =...

Given,

$\left(\alpha, \alpha\right)$ lies on $C: \left({x}^{2} + {y}^{2} - 3\right) + {\left({x}^{2} - {y}^{2} - 1\right)}$^{5} = 0$</p><p>So on putting \left(\alpha, \alpha\right)$ we get,

$2{\alpha}^{2} - 3 - {1}^{5} = 0$

$⇒ \alpha = \sqrt{2}$

Now, differentiating the curve $C$ we get,

$2x + 2y \cdot {y}^{'} + 5{\left({x}^{2} - {y}^{2} - 1\right)}^{4}\left(2x - 2y{y}^{'}\right) = 0 ⋯ \left(1\right)$

At \left(\sqrt{2}, \sqrt{2}\right)$</p><p>$\sqrt{2} + \sqrt{2}{y}^{'} + 5{\left(- 1\right)}^{4}$\left(\sqrt{2} - \sqrt{2}{y}^{'}\right) = 0$

$⇒ {y}^{'} = \dfrac{3}{2} ⋯ \left(2\right)$

Again, Diff. \left(1\right)$w.r.t.$x$we get,</p><p>$1 + {\left({y}^{'}\right)}^{2} + yy{y}^{' '} + 20{\left({x}^{2} - {y}^{2} - 1\right)}^{3}{\left(x - y{y}^{'}\right)}^{2} \cdot 2

+ 5{\left({x}^{2} - {y}^{2} - 1\right)}^{4}\left(1 - {\left({y}^{'}\right)}^{2} - yy{y}^{' '}\right) = 0

At \left(\sqrt{2}, \sqrt{2}\right)$and${y}^{'} = \dfrac{3}{2}</p><p>We have,</p><p>\left(1 + \dfrac{9}{4}\right) + \sqrt{2}{y}^{' '} - 40{\left(\sqrt{2} - \sqrt{2} \cdot \dfrac{3}{2}\right)}^{2}

+ 5\left(1\right)\left(1 - \dfrac{9}{4} - \sqrt{2}{y}^{' '}\right) = 0

⇒ 4\sqrt{2}{y}^{' '} = - 23

{y}^{' '} = \dfrac{- 23}{4\sqrt{2}}

So, value of 3{y}^{'} - {y}^{3}{y}^{' '} = \dfrac{9}{2} + \dfrac{23}{2} = 16