$\frac{dy}{dx}\ = \frac{-1}{x^3}\\$
is the solution:
$y = \frac{1}{2x^2} + k$
or
$y = \frac{1}{2x^2} - k$
Both are correct, with the understanding that $k$ is an arbitrary constant.
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Both are correct, with the understanding that $k$ is an arbitrary constant.