Given that the slope of the tangent to a curve $$y = y(x)$$ at any point $$(x, y)$$ is $$\frac{2y}{x^{2}}$$. If the curve passes through the centre of the circle $$x^{2} + y^{2} - 2x - 2y = 0$$, then its equation is:

We are told that the slope of the tangent to the required curve is given by the derivative formula

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

To find the equation of the curve, we first separate the variables. Bringing all the terms involving $$y$$ to one side and those involving $$x$$ to the other, we have

$$\frac{1}{y}\,dy = \frac{2}{x^{2}}\,dx.$$

Now we integrate both sides. The standard integral formulas to be used are:

$$\int \frac{1}{y}\,dy = \ln|y| + C_{1}, \quad \int x^{-2}\,dx = -x^{-1} + C_{2}.$$

Applying these, we obtain

$$\int \frac{1}{y}\,dy = \int \frac{2}{x^{2}}\,dx,$$

so

$$\ln|y| = 2\int x^{-2}\,dx = 2\left(-x^{-1}\right) + C,$$

which simplifies to

$$\ln|y| = -\frac{2}{x} + C.$$

Here $$C$$ is the constant of integration. We next determine this constant by using the given condition that the curve passes through the centre of the circle

$$x^{2} + y^{2} - 2x - 2y = 0.$$

To find the centre, we complete the square:

$$x^{2} - 2x + y^{2} - 2y = 0 \;\;\Longrightarrow\;\; (x - 1)^{2} - 1 + (y - 1)^{2} - 1 = 0 \;\;\Longrightarrow\;\; (x - 1)^{2} + (y - 1)^{2} = 2.$$

Hence the centre is $$\bigl(1,\;1\bigr).$$ Because the required curve passes through this point, we substitute $$x = 1,\; y = 1$$ into the relation $$\ln|y| = -\dfrac{2}{x} + C$$:

$$\ln|1| = -\frac{2}{1} + C.$$

Since $$\ln|1| = 0$$, we get

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

Therefore the equation obtained after integration becomes

$$\ln|y| = 2 - \frac{2}{x}.$$

We now multiply both sides by $$x$$ to clear the denominator:

$$x\,\ln|y| = 2(x - 1).$$

This matches Option B:

$$x \log_e|y| = 2(x - 1).$$

Hence, the correct answer is Option B.