Step-by-step explanation:

We have,

[tex](x^{2}y) \dfrac{dy}{dx} -xy^{2} =1 \\ [/tex]

Taking differentials,

[tex] \implies \: x^{2}y \: dy-xy^{2} \: dx =dx \\ [/tex]

[tex] \implies \: xy \left(x\: dy-y \: dx \right) =dx \\ [/tex]

[tex] \implies \: \dfrac{xy \left(x\: dy-y \: dx \right)}{ {x}^{2} } = \dfrac{dx}{ {x}^{2} } \\ [/tex]

[tex] \implies \: xy \left(\dfrac{x\: dy-y \: dx }{ {x}^{2} } \right) = \dfrac{dx}{ {x}^{2} } \\ [/tex]

Now,

[tex] \bf{Let \: \: f = \dfrac{y}{x} }[/tex]

[tex] \bf{ \mapsto\: \: \dfrac{df }{dx}= \dfrac{x \dfrac{dy}{dx} - y}{ {x}^{2} } }[/tex]

[tex] \bf{ \mapsto\: \: df = \dfrac{x \: dy- y \: dx}{ {x}^{2} } }[/tex]

[tex] \bf{ \mapsto\: \: d \left( \dfrac{y}{x} \right) = \dfrac{x \: dy- y \: dx}{ {x}^{2} } }[/tex]

So,

[tex] \implies \: xy \: d\left(\dfrac{y}{x} \right) = \dfrac{dx}{ {x}^{2} } \\ [/tex]

[tex] \implies \: \dfrac{xy}{ {x}^{2} }\: d\left(\dfrac{y}{x} \right) = \dfrac{dx}{ {x}^{4} } \\ [/tex]

[tex] \implies \: \dfrac{y}{ x }\: d\left(\dfrac{y}{x} \right) = \dfrac{dx}{ {x}^{4} } \\ [/tex]

Integrating both sides,

[tex] \displaystyle \: \implies \: \int\dfrac{y}{ x }\: d\left(\dfrac{y}{x} \right) = \int\dfrac{dx}{ {x}^{4} } \\ [/tex]

[tex] \displaystyle \: \implies \: \dfrac{1}{2} \left(\dfrac{y}{ x }\right)^{2} = \dfrac{1}{3 {x}^{3} } + c \\ [/tex]

[tex] \displaystyle \: \implies \: \dfrac{ {y}^{2} }{ 2{x}^{2} } = \dfrac{1}{3 {x}^{3} } + c \\ [/tex]

Multiply both sides by 6x³

[tex] \displaystyle \: \implies \: 3x {y}^{2} = 2+ 6 c{x}^{3} \\ [/tex]

Put 6c = C,

[tex] \displaystyle \: \implies \: 3x {y}^{2} = 2+C {x}^{3} \\ [/tex]