The slope of normal at any point $$(x, y), x > 0, y > 0$$ on the curve $$y = y(x)$$ is given by $$\frac{x^2}{xy - x^2y^2 - 1}$$. If the curve passes through the point $$(1, 1)$$, then $$e \cdot y(e)$$ is equal to

The slope of the normal at any point $$(x, y)$$ on the curve is given by $$\frac{x^2}{xy - x^2y^2 - 1}$$.

Since the slope of the normal is $$-\frac{1}{dy/dx}$$, we have:

$$-\frac{1}{dy/dx} = \frac{x^2}{xy - x^2y^2 - 1}$$

$$\frac{dy}{dx} = \frac{x^2y^2 - xy + 1}{x^2} = y^2 - \frac{y}{x} + \frac{1}{x^2}$$

Let $$v = xy$$, so $$y = \frac{v}{x}$$ and $$\frac{dy}{dx} = \frac{v'x - v}{x^2}$$, where $$v' = \frac{dv}{dx}$$.

Substituting:

$$\frac{v'x - v}{x^2} = \frac{v^2}{x^2} - \frac{v}{x^2} + \frac{1}{x^2}$$

Multiplying both sides by $$x^2$$:

$$v'x - v = v^2 - v + 1$$

$$v'x = v^2 + 1$$

Separating variables:

$$\frac{dv}{v^2 + 1} = \frac{dx}{x}$$

Integrating both sides:

$$\tan^{-1}(v) = \ln|x| + C$$

$$\tan^{-1}(xy) = \ln x + C$$

Using the initial condition $$(1, 1)$$:

$$\tan^{-1}(1 \cdot 1) = \ln 1 + C$$

$$\frac{\pi}{4} = 0 + C$$, so $$C = \frac{\pi}{4}$$.

The solution is: $$\tan^{-1}(xy) = \ln x + \frac{\pi}{4}$$

At $$x = e$$:

$$\tan^{-1}(e \cdot y(e)) = \ln e + \frac{\pi}{4} = 1 + \frac{\pi}{4}$$

$$e \cdot y(e) = \tan\left(1 + \frac{\pi}{4}\right)$$

Using the addition formula $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$:

$$\tan\left(1 + \frac{\pi}{4}\right) = \frac{\tan 1 + \tan\frac{\pi}{4}}{1 - \tan 1 \cdot \tan\frac{\pi}{4}} = \frac{\tan 1 + 1}{1 - \tan 1} = \frac{1 + \tan 1}{1 - \tan 1}$$

Therefore: $$e \cdot y(e) = \frac{1 + \tan(1)}{1 - \tan(1)}$$

The correct answer is Option D.