What is the solution for y^'= 1/x y-((x^2)/3)y^4 ?

The solution for y^'= 1/x y-((x^2)/3)y^4 is y=(\sqrt[3]{6}x)/(\sqrt[3]{x^6+c_{1)}}

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y^'= 1/x y-((x^2)/3)y^4

y^{^{'}} = \frac{1}{x} y - (\frac{x ^{2}}{3}) y^{4}

Solution steps

y^{'} (x) = \frac{1}{x} y - (\frac{x ^{2}}{3}) y^{4}

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What is the solution for y^'= 1/x y-((x^2)/3)y^4 ?
The solution for y^'= 1/x y-((x^2)/3)y^4 is y=(\sqrt[3]{6}x)/(\sqrt[3]{x^6+c_{1)}}