If x 3 - 2 x 2 y 2 + 5 x + y - 5 = 0 , then at 1 , 1 ,   y ' ' 1 =

Given,

$x^3-2x^2{y}^2+5x+y-5=0$

Now derivative of equation w.r.t. $x$ will be

$\Rightarrow 3{\mathrm{x}}^2-4{\mathrm{xy}}^2-4{\mathrm{x}}^2{\mathrm{yy}}^{\text{'}}+5+{\mathrm{y}}^{\text{'}}-0=0 ...\left(1\right)$

Given that $y\left(1\right)=1$

$\Rightarrow 3-4-4y^{\text{'}}\left(1\right)+5+y^{\text{'}}\left(1\right)=0\Rightarrow y^{\text{'}}\left(1\right)=\frac{4}{3}$

Derivative of equation $\left(1\right)$ w.r.t. $x$ will be

$\Rightarrow 6x-4y^2-8xy{y}^{\text{'}}-8xy{y}^{\text{'}}-4x^2{y}^{\text{'}}{y}^{\text{'}}-4x^{2}y{y}^{\text{'}\text{'}}+y^{\text{'}\text{'}}=0$

$\Rightarrow 6-4-8\times \frac{4}{3}-8\times \frac{4}{3}-4\times \frac{16}{9}-4y^{\text{'}\text{'}}\left(1\right)+y^{\text{'}\text{'}}\left(1\right)=0$

$\Rightarrow y^{\text{'}\text{'}}\left(1\right)=\frac{-238}{27}$