Question:

Step 1: Rewrite the differential equation in standard form. We start with the given differential equation: $$y^2\,dx + (x^2 - xy + y^2)\,dy = 0$$ Rearranging to express $\frac{dx}{dy}$: $$y^2\,dx = -(x^2 - xy + y^2)\,dy$$ $$\frac{dx}{dy} = -\frac{x^2 - xy + y^2}{y^2}$$ Separating the terms: $$\frac{dx}{dy} = -\frac{x^2}{y^2} + \frac{x}{y} - 1$$ Step 2: Apply the substitution $v = \frac{x}{y}$ to convert to a separable equation. Let $v = \frac{x}{y}$, which means $x = vy$. Taking the derivative with respect to $y$: $$\frac{dx}{dy} = v + y\frac{dv}{dy}$$ Substituting into our equation from Step 1: $$v + y\frac{dv}{dy} = -v^2 + v - 1$$ Simplifying: $$y\frac{dv}{dy} = -v^2 - 1$$ Separating variables: $$\frac{dv}{v^2+1} = -\frac{dy}{y}$$ Step 3: Integrate both sides of the separated equation. Integrating the left side with respect to $v$ and the right side with respect to $y$: $$\int \frac{dv}{v^2+1} = -\int \frac{dy}{y}$$ $$\tan^{-1}(v) = -\ln y + C_1$$ Step 4: Substitute back $v = \frac{x}{y}$ and rearrange to obtain the final solution. $$\tan^{-1}\left(\frac{x}{y}\right) = -\ln y + C_1$$ $$\tan^{-1}\left(\frac{x}{y}\right) + \ln y = C_1$$ Rewriting with $C = -C_1$: $$\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0$$ However, examining the given options more carefully, we note that Option 3 is the correct answer provided. The solution $\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0$ matches the form we derived. **Final Answer:** The solution of the differential equation is: $$\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0$$ This corresponds to **Option 3**.