Answer
$y=|x|\sqrt Cx-1$
Work Step by Step
We are given:
$$(3y^2+x^2)dx-2xydy=0$$
We have that:
$$M(x,y)=3y^2+x^2\wedge N(x,y)=-2xy$$
then
$$M'(x,y)=6y$$
$$N'(x,y)=-2y$$
So this equation is not exist.
But we have:
$$\frac{M'_y-N'_y}{M}=\frac{8y}{-2xy}=-\frac{4}{x}$$
Integrating factor:
$$I=e^{-\int -\frac{4}{x}dx}=e^{-4\ln x}=x^{-4}$$
Multiplying the equation by $x^{-4}$
$$(3y^2x^{-4}+x^2)dx-2x^{-3}ydy=0$$
There exists a potential function $\phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
$$\frac{\partial \phi}{\partial x}=3y^2x^{-4}+x^2 \wedge \frac{\partial \phi}{\partial y}=-2x^{-3}y$$
Integrating $\frac{\partial \phi}{\partial x}=3y^2x^{-4}+x^2$ we get
$$\phi=-y^2x^{-3}-x^{-1}+f(y)$$
Since $\frac{\partial \phi}{\partial y}=-2yx^{-3}+f'(y)$
We obtain:
$$f'(y)=0\rightarrow f(y)=C$$
$C$ is constant of integration
The potential solution is:
$$\phi=-y^2x^{-3}+x^{-1}$$
$$y^2=Cx^3-x^2$$
Hence general solution is $y=|x|\sqrt Cx-1$
