Official solution
\[{{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)\]
\[ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}\]
\[ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}\]
\[ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}\]
\[ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}\]