What is the solution for (dy)/(dx)= 1/(3x^2y^2) ?

The solution for (dy)/(dx)= 1/(3x^2y^2) is y=\sqrt[3]{-1/x+c_{1}}

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(dy)/(dx)= 1/(3x^2y^2)

\frac{d y}{d x} = \frac{1}{3 x ^{2} y ^{2}}

Solution steps

\frac{d y}{d x} = \frac{1}{3 x ^{2} y ^{2}}

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What is the solution for (dy)/(dx)= 1/(3x^2y^2) ?
The solution for (dy)/(dx)= 1/(3x^2y^2) is y=\sqrt[3]{-1/x+c_{1}}