(2)(x^3+y^3)dx-3xy^2dy=0.

将方程改写为 $\frac{dy}{dx} = \frac{x^3 + y^3}{3xy^2}$。令 $u = \frac{y}{x}$，则 $y = ux$，$\frac{dy}{dx} = u + x\frac{du}{dx}$。代入得：

$u + x\frac{du}{dx} = \frac{1 + u^3}{3u^2} \implies x\frac{du}{dx} = \frac{1 - 2u^3}{3u^2}$

分离变量并积分：

$\int \frac{3u^2}{1 - 2u^3} du = \int \frac{dx}{x} \implies -\frac{1}{2} \ln |1 - 2u^3| = \ln |x| + C_1$

解得：

$|1 - 2u^3| = \frac{C}{x^2} \implies 1 - 2\left(\frac{y}{x}\right)^3 = \frac{C}{x^2} \implies x^3 - 2y^3 = Cx$

答案： $\boxed{x^3 - 2y^3 = Cx}$