1. **State the problem:** Solve the differential equation $$(x^2 + y^2) \, dx - 2xy \, dy = 0.$$\n\n2. **Rewrite the equation:** We can write it as $$(x^2 + y^2) + (-2xy) \frac{dy}{dx} = 0,$$ which implies $$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}.$$\n\n3. **Use substitution:** Let $$v = \frac{y}{x} \Rightarrow y = vx.$$ Then $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ by the product rule.\n\n4. **Substitute into the differential equation:**\n$$v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2 x^2}{2 v x^2} = \frac{1 + v^2}{2v}.$$\n\n5. **Isolate $\frac{dv}{dx}$:**\n$$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}.$$\n\n6. **Separate variables:**\n$$\frac{2v}{1 - v^2} dv = \frac{dx}{x}.$$\n\n7. **Integrate both sides:**\n$$\int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x}.$$\n\n8. **Simplify the integral on the left:**\nRewrite denominator: $$1 - v^2 = (1 - v)(1 + v).$$\nUse substitution or partial fractions: Let’s set $$u = 1 - v^2 \Rightarrow du = -2v dv,$$ so $$-du = 2v dv.$$\nThus,\n$$\int \frac{2v}{1 - v^2} dv = \int \frac{-du}{u} = -\ln|u| + C = -\ln|1 - v^2| + C.$$\n\n9. **Integrate right side:**\n$$\int \frac{dx}{x} = \ln|x| + C.$$\n\n10. **Combine results:**\n$$-\ln|1 - v^2| = \ln|x| + C_1,$$ or equivalently,\n$$\ln|x| + \ln|1 - v^2| = C_2,$$ where $$C_2 = -C_1.$$\n\n11. **Rewrite in terms of $x$ and $y$:**\nRecall $$v = \frac{y}{x},$$ so\n$$\ln|x(1 - (\frac{y}{x})^2)| = C_2,$$\nwhich simplifies to\n$$\ln|x - \frac{y^2}{x}| = C_2,$$\n$$\ln\left|\frac{x^2 - y^2}{x}\right| = C_2.$$\n\n12. **Exponentiate both sides:**\n$$\left|\frac{x^2 - y^2}{x}\right| = K,$$ where $$K = e^{C_2} > 0.$$\n\n13. **Final implicit solution:**\n$$|x^2 - y^2| = K |x|,$$ or equivalently,\n$$x^2 - y^2 = C x,$$ where $$C$$ is an arbitrary constant (absorbing sign into $$C$$).