${{{d^2}x} \over {d{y^2}}}$ equals:

Question

$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$

${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$

$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$

${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$

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