$$\because$$ $$y(x) = {\left( {{x^x}} \right)^x}$$

$$\therefore$$ $$y = {x^{{x^2}}}$$

$$\therefore$$ $${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$$

$$\therefore$$ $${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$$ ..... (i)

Now, $${{{d^2}x} \over {dx^2}} = {d \over {dx}}\left( {{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 1}}} \right)\,.\,{{dx} \over {dy}}$$

$$ = {{ - x{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 2}}\,.\,{x^{{x^2}}}(1 + 2\ln x)({x^2} + 2{x^2}\ln x + 3)} \over {{x^{{x^2}}}(1 + 2\ln x)}}$$

$$ = {{ - {x^{{x^2}}}(1 + 2\ln x)({x^3} + 3 + 2{x^2}\ln x)} \over {{{\left( {{x^{{x^2}}}(1 + 2\ln x)} \right)}^3}}}$$

$${{{d^2}x} \over {d{y^2}(at\,x = 1)}} = - 4$$

$$\therefore$$ $${{{d^2}x} \over {d{y^2}(at\,x = 1)}} + 20 = 16$$