** Final answer to the problem

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** Step-by-step Solution **

** How should I solve this problem?

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Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

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The differential equation $3y^2dy-2xdx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$

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Using the test for exactness, we check that the differential equation is exact

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Integrate $M(x,y)$ with respect to $x$ to get

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Now take the partial derivative of $-x^2$ with respect to $y$ to get

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Set $3y^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

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Find $g(y)$ integrating both sides

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We have found our $f(x,y)$ and it equals

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Then, the solution to the differential equation is

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Find the explicit solution to the differential equation. We need to isolate the variable $y$

** Final answer to the problem

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