The solution of the differential equation, $$\frac{dy}{dx} = (x - y)^2$$, when $$y(1) = 1$$, is:

We start with the given differential equation

$$\frac{dy}{dx} = (x - y)^2.$$

Because the right-hand side contains the combination $$x-y$$, it is natural to introduce the substitution

$$v = x - y.$$

From this we have $$y = x - v$$, and on differentiating both sides with respect to $$x$$ we obtain

$$\frac{dy}{dx} = 1 - \frac{dv}{dx}.$$

Now we substitute $$y = x - v$$ and $$\dfrac{dy}{dx} = 1 - \dfrac{dv}{dx}$$ into the original equation. We get

$$1 - \frac{dv}{dx} = v^2.$$

Rearranging, we obtain

$$\frac{dv}{dx} = 1 - v^2.$$

Next, we separate the variables. Writing all terms involving $$v$$ on the left and all terms involving $$x$$ on the right, we get

$$\frac{dv}{1 - v^2} = dx.$$

To integrate we recall the standard integral formula

$$\int \frac{dv}{1 - v^2} = \frac12 \ln\left|\frac{1 + v}{1 - v}\right| + C,$$

where $$C$$ is the constant of integration. Using this result, integrating both sides gives

$$\frac12 \ln\left|\frac{1 + v}{1 - v}\right| = x + C_1.$$

Multiplying by $$2$$ we have

$$\ln\left|\frac{1 + v}{1 - v}\right| = 2x + C_2,$$

where $$C_2 = 2C_1$$ is another constant of integration.

At this stage we impose the initial condition. The problem states that $$y(1) = 1$$, and since $$v = x - y$$, at $$x = 1$$ we have

$$v(1) = 1 - 1 = 0.$$

Substituting $$x = 1$$ and $$v = 0$$ in the integrated equation gives

$$\ln\left|\frac{1 + 0}{1 - 0}\right| = 2(1) + C_2.$$

The logarithm on the left simplifies to $$\ln 1 = 0$$, so

$$0 = 2 + C_2 \quad\Longrightarrow\quad C_2 = -2.$$

Replacing $$C_2$$ by $$-2$$ in our integrated equation we get

$$\ln\left|\frac{1 + v}{1 - v}\right| = 2x - 2 = 2(x - 1).$$

Finally we return to the original variables by recalling that $$v = x - y$$. Substituting gives

$$\ln\left|\frac{1 + x - y}{1 - x + y}\right| = 2(x - 1).$$

Using the logarithmic identity $$\ln\left|\dfrac{a}{b}\right| = -\ln\left|\dfrac{b}{a}\right|$$, we can equivalently write the result as

$$-\ln\left|\frac{1 - x + y}{1 + x - y}\right| = 2(x - 1).$$

This matches exactly with Option B.

Hence, the correct answer is Option B.