Let $$y = y(x)$$ be the solution curve of the differential equation $$\frac{dy}{dx} = \frac{y}{x}(1 - xy^2(1 + \log_e x))$$, $$x \gt 0$$, $$y(1) = 3$$. Then $$\frac{y^2(x)}{9}$$ is equal to:

We consider the differential equation $$\frac{dy}{dx} = \frac{y}{x}\left(1 - xy^2(1 + \log_e x)\right), \quad x \gt 0, \quad y(1) = 3$$. Since rewriting the right-hand side gives $$\frac{dy}{dx} = \frac{y}{x} - (1 + \ln x)\,y^3$$, this is a Bernoulli equation with exponent $$n = 3$$. Substituting $$v = y^{-2}$$ and noting that $$\frac{dv}{dx} = -2y^{-3}\frac{dy}{dx}$$ leads us to divide the original ODE by $$y^3$$, yielding $$y^{-3}\frac{dy}{dx} = \frac{y^{-2}}{x} - (1 + \ln x)$$. Therefore, we have $$-\frac{1}{2}\frac{dv}{dx} = \frac{v}{x} - (1 + \ln x)$$ or equivalently $$\frac{dv}{dx} + \frac{2v}{x} = 2(1 + \ln x)\,.$$

The integrating factor is $$e^{\int \frac{2}{x}dx} = x^2$$, so multiplying through gives $$\frac{d}{dx}(x^2 v) = 2x^2(1 + \ln x)\,. $$ Integrating the right-hand side produces $$\int 2x^2(1 + \ln x)\,dx = 2\left[\frac{x^3}{3} + \frac{x^3 \ln x}{3} - \frac{x^3}{9}\right] = \frac{2x^3}{9}(2 + 3\ln x)$$ and hence $$x^2 v = \frac{2x^3(2 + 3\ln x)}{9} + C\,. $$

Applying the initial condition $$y(1) = 3$$ implies $$v(1) = \frac{1}{9}$$, which gives $$1 \cdot \frac{1}{9} = \frac{2(2)}{9} + C \implies C = \frac{1}{9} - \frac{4}{9} = -\frac{3}{9} = -\frac{1}{3}\,. $$ Therefore $$x^2 v = \frac{2x^3(2 + 3\ln x)}{9} - \frac{1}{3} = \frac{2x^3(2 + 3\ln x) - 3}{9}\,. $$ Since $$v = \frac{1}{y^2}$$ it follows that $$\frac{x^2}{y^2} = \frac{2x^3(2 + 3\ln x) - 3}{9}$$ or equivalently $$\frac{y^2}{9} = \frac{x^2}{2x^3(2 + 3\ln x) - 3}\,. $$ Using $$3\ln x = \ln x^3$$ yields $$\frac{y^2(x)}{9} = \frac{x^2}{2x^3(2 + \log_e x^3) - 3}\,, $$ which corresponds to Option B.