Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) - 1$$, $$y(0) = 1$$. Then, $$\left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2$$ equals:

We need to solve the differential equation $$\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) - 1$$ with $$y(0) = 1$$ and find $$\left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2$$.

We set $$v = x + y$$ so that $$\frac{dv}{dx} = 1 + \frac{dy}{dx}$$, which gives $$\frac{dy}{dx} = \frac{dv}{dx} - 1$$. Substituting into the differential equation yields $$\frac{dv}{dx} - 1 = 2xv^3 - xv - 1$$, which simplifies to $$\frac{dv}{dx} = 2xv^3 - xv = xv(2v^2 - 1).$$

Separating variables leads to $$\frac{dv}{v(2v^2 - 1)} = x\,dx.$$

We perform partial fraction decomposition by writing $$\frac{1}{v(2v^2 - 1)} = \frac{A}{v} + \frac{Bv + C}{2v^2 - 1}.$$ Clearing denominators gives $$1 = A(2v^2 - 1) + (Bv + C)v = 2Av^2 - A + Bv^2 + Cv.$$ Comparing coefficients for $$v^2$$, $$v^1$$, and the constant term yields $$2A + B = 0$$, $$C = 0$$, and $$-A = 1$$, so $$A = -1$$, $$B = 2$$, and $$C = 0$$. Therefore $$\frac{1}{v(2v^2 - 1)} = \frac{-1}{v} + \frac{2v}{2v^2 - 1}.$$

Integrating both sides gives $$\int\left(\frac{-1}{v} + \frac{2v}{2v^2 - 1}\right)dv = \int x\,dx$$, which leads to $$-\ln|v| + \tfrac{1}{2}\ln|2v^2 - 1| = \tfrac{x^2}{2} + C_0$$, or equivalently $$\ln\left|\frac{\sqrt{2v^2 - 1}}{v}\right| = \tfrac{x^2}{2} + C_0$$. Exponentiating yields $$\frac{\sqrt{2v^2 - 1}}{v} = K e^{x^2/2}$$, where $$K = \pm e^{C_0}$$.

Applying the initial condition $$y(0) = 1$$ gives $$v(0) = x + y = 1$$, so $$\frac{\sqrt{2(1)^2 - 1}}{1} = K$$ and hence $$K = 1$$. It follows that $$\frac{\sqrt{2v^2 - 1}}{v} = e^{x^2/2}$$.

To find the desired expression, note that $$\left(\frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right)\right)^2 = v\left(\tfrac{1}{\sqrt{2}}\right)^2$$, where $$v_0 = v\bigl(\tfrac{1}{\sqrt{2}}\bigr)$$. From the solution we have $$\frac{\sqrt{2v_0^2 - 1}}{v_0} = e^{1/4}$$. Squaring both sides gives $$\frac{2v_0^2 - 1}{v_0^2} = e^{1/2} = \sqrt{e}$$, so $$2 - \frac{1}{v_0^2} = \sqrt{e}$$. Hence $$\frac{1}{v_0^2} = 2 - \sqrt{e}$$ and $$v_0^2 = \frac{1}{2 - \sqrt{e}}$$. Therefore the required value is $$\frac{1}{2 - \sqrt{e}}$$, which corresponds to Option D.