Let $$f(x) = \sqrt{\lim_{r \to x}\left\{\frac{2r^2[(f(r))^2 - f(x)f(r)]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}}\right\}}$$ be differentiable in $$(-\infty, 0) \cup (0, \infty)$$ and $$f(1) = 1$$. Then the value of $$ae$$, such that $$f(a) = 0$$, is equal to ______.

Squaring the given limit definition equation:

$$(f(x))^2 = \frac{2x^2 f(x) f'(x)}{2x} - x^3 e^{\frac{f(x)}{x}} \implies (f(x))^2 = x f(x) f'(x) - x^3 e^{\frac{f(x)}{x}}$$

Divide by $$x^2$$ and substitute $$f(x) = vx$$:

$$v^2 = v\left(v + x\frac{dv}{dx}\right) - xe^v \implies v e^{-v} dv = dx$$

Integrating both sides:

$$-e^{-v}(v+1) = x + C$$

Using $$f(1) = 1 \implies v = 1$$ at $$x = 1$$:

$$-e^{-1}(2) = 1 + C \implies C = -1 - \frac{2}{e}$$

 $$f(a) = 0 \implies v = 0$$:

$$-e^0(0+1) = a - 1 - \frac{2}{e} \implies -1 = a - 1 - \frac{2}{e} \implies a = \frac{2}{e}$$

$$ae = \left(\frac{2}{e}\right)e = 2$$.