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The given equation is $(x^{3}-3xy^{2})dx=(y^{3}-3x^{2}y)dy$. We can write this as: $$\frac{dy}{dx} = \frac{x^{3}-3xy^{2}}{y^{3}-3x^{2}y}$$ This is a homogeneous differential equation of degree 3.
Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these into the equation: $$v + x\frac{dv}{dx} = \frac{x^{3}-3x(vx)^{2}}{(vx)^{3}-3x^{2}(vx)} = \frac{x^{3}(1-3v^{2})}{x^{3}(v^{3}-3v)} = \frac{1-3v^{2}}{v^{3}-3v}$$
Subtract $v$ from both sides: $$x\frac{dv}{dx} = \frac{1-3v^{2}}{v^{3}-3v} - v = \frac{1-3v^{2}-v^{4}+3v^{2}}{v^{3}-3v} = \frac{1-v^{4}}{v^{3}-3v}$$ Rearranging for integration: $$\int \frac{v^{3}-3v}{1-v^{4}} dv = \int \frac{1}{x} dx$$
Split the integral: $$\int \frac{v^{3}}{1-v^{4}} dv - 3\int \frac{v}{1-v^{4}} dv = \ln|x| + C$$ Using substitution $u = 1-v^{4}$, $du = -4v^{3}dv$, the first part integrates to $-\frac{1}{4}\ln|1-v^{4}|$. The second part involves partial fractions or standard forms. Solving leads to: $$(x^{2}+y^{2})^{2} = C(x^{2}-y^{2})$$
Final Answer: (x^{2}+y^{2})^{2} = C(x^{2}-y^{2})