$\because$ $y(x) = {\left( {{x^x}} \right)^x}$ $\therefore$ $y = {x^{{x^2}}}$ $\therefore$ ${{dy} \over {dx}} = {x^2}\,.\,{x^{{x^2} - 1}} + {x^{{x^2}}}\ln x\,.\,2x$ $\therefore$ ${{dx} \over {dy}} = {1 \over {{x^{{x^2} + 1}}(1 + 2\ln x)}}$ ..... (i) Now, ${{{d^2}x} \over {dx^2}} = {d \over {dx}}\left( {{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 1}}} \right)\,.\,{{dx} \over {dy}}$ $= {{ - x{{\left( {{x^{{x^2} + 1}}(1 + 2\ln x)} \right)}^{ - 2}}\,.\,{x^{{x^2}}}(1 + 2\ln x)({x^2} + 2{x^2}\ln x + 3)} \over {{x^{{x^2}}}(1 + 2\ln x)}}$ $= {{ - {x^{{x^2}}}(1 + 2\ln x)({x^3} + 3 + 2{x^2}\ln x)} \over {{{\left( {{x^{{x^2}}}(1 + 2\ln x)} \right)}^3}}}$ ${{{d^2}x} \over {d{y^2}(at\,x = 1)}} = - 4$ $\therefore$ ${{{d^2}x} \over {d{y^2}(at\,x = 1)}} + 20 = 16$