If $y={\left(x^x\right)}^x\text{\hspace{0.17em},\hspace{0.17em}}x>0$ , then the value of  $\frac{d^{2}x}{d{y}^2}+20$ at x = 1 is equal to

Given function is  $y\left(x\right)={\left(x^x\right)}^x=x^{x^2}$
Take log both sides,  $\mathcal{l}n\left(x\right)=x^2\cdot \mathcal{l}nx$
Differentiate w.r.t x both sides,
$\frac{1}{y\left(x\right)}\cdot y^1\left(x\right)=\frac{x^2}{x}+2x\cdot \mathcal{l}nx$
$y^1\left(x\right)=y\left(x\right)[x+2x\mathcal{l}nx]$
Put x = 1 in above equation
$y\left(1\right)=1;y^1\left(1\right)=1$
$y^{11}\left(x\right)=y^1\left(x\right)[x+2x\cdot \mathcal{l}n(x\left)\right]+y\left(x\right)[1+2(1+\mathcal{l}nx\left)\right]$
$y^{11}\left(1\right)=1(1+0)+1(1+2)=4$
$\frac{d^{2}y}{d{x}^2}=-{\left(\frac{dy}{dx}\right)}^2\cdot \frac{d^{2}x}{d{y}^2}\Rightarrow 4=-\frac{d^{2}y}{d{x}^2}$
$\frac{d^{2}x}{d{y}^2}=4\Rightarrow -4+20=16$