If $$f(x) = \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x - 1$$, $$x \in R$$, then the equation $$f(x) = 0$$ has:

We are given the function $$ f(x) = \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x - 1 $$ for all real numbers $$ x $$, and we need to solve $$ f(x) = 0 $$. This means we must find the values of $$ x $$ such that $$ \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1 $$.

First, note that both $$ \frac{3}{5} $$ and $$ \frac{4}{5} $$ are fractions less than 1, so as $$ x $$ increases, each term $$ \left(\frac{3}{5}\right)^x $$ and $$ \left(\frac{4}{5}\right)^x $$ decreases exponentially. To understand the behavior of $$ f(x) $$, we evaluate it at some specific points.

At $$ x = 0 $$:
$$ f(0) = \left(\frac{3}{5}\right)^0 + \left(\frac{4}{5}\right)^0 - 1 = 1 + 1 - 1 = 1 $$, which is greater than 0.

At $$ x = 1 $$:
$$ f(1) = \frac{3}{5} + \frac{4}{5} - 1 = \frac{7}{5} - 1 = \frac{2}{5} $$, which is greater than 0.

At $$ x = 2 $$:
$$ f(2) = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 - 1 = \frac{9}{25} + \frac{16}{25} - 1 = \frac{25}{25} - 1 = 1 - 1 = 0 $$. So, $$ x = 2 $$ is a solution.

At $$ x = 3 $$:
$$ f(3) = \left(\frac{3}{5}\right)^3 + \left(\frac{4}{5}\right)^3 - 1 = \frac{27}{125} + \frac{64}{125} - 1 = \frac{91}{125} - 1 = \frac{91 - 125}{125} = \frac{-34}{125} $$, which is less than 0.

Now, check for negative values. At $$ x = -1 $$:
$$ f(-1) = \left(\frac{3}{5}\right)^{-1} + \left(\frac{4}{5}\right)^{-1} - 1 = \frac{5}{3} + \frac{5}{4} - 1 $$.
Compute $$ \frac{5}{3} + \frac{5}{4} = \frac{20}{12} + \frac{15}{12} = \frac{35}{12} $$, then $$ f(-1) = \frac{35}{12} - 1 = \frac{35}{12} - \frac{12}{12} = \frac{23}{12} $$, which is greater than 0.

At $$ x = -2 $$:
$$ f(-2) = \left(\frac{3}{5}\right)^{-2} + \left(\frac{4}{5}\right)^{-2} - 1 = \left(\frac{5}{3}\right)^2 + \left(\frac{5}{4}\right)^2 - 1 = \frac{25}{9} + \frac{25}{16} - 1 $$.
Compute $$ \frac{25}{9} + \frac{25}{16} = 25 \left( \frac{1}{9} + \frac{1}{16} \right) = 25 \left( \frac{16 + 9}{144} \right) = 25 \times \frac{25}{144} = \frac{625}{144} $$, then $$ f(-2) = \frac{625}{144} - 1 = \frac{625}{144} - \frac{144}{144} = \frac{481}{144} $$, which is greater than 0.

We have found that $$ f(2) = 0 $$, and for $$ x < 2 $$ (like $$ x = -2, -1, 0, 1 $$), $$ f(x) > 0 $$, and for $$ x > 2 $$ (like $$ x = 3 $$), $$ f(x) < 0 $$. To determine if there are more solutions, we analyze the derivative to see if the function is monotonic.

Define $$ g(x) = \left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x $$, so $$ f(x) = g(x) - 1 $$. The derivative of $$ g(x) $$ is:
$$ g'(x) = \frac{d}{dx} \left[ \left(\frac{3}{5}\right)^x \right] + \frac{d}{dx} \left[ \left(\frac{4}{5}\right)^x \right] $$.
Since $$ \frac{d}{dx} a^x = a^x \ln a $$, we have:
$$ g'(x) = \left(\frac{3}{5}\right)^x \ln \left(\frac{3}{5}\right) + \left(\frac{4}{5}\right)^x \ln \left(\frac{4}{5}\right) $$.

Now, $$ \ln \left(\frac{3}{5}\right) < 0 $$ and $$ \ln \left(\frac{4}{5}\right) < 0 $$ because both fractions are less than 1. Also, $$ \left(\frac{3}{5}\right)^x > 0 $$ and $$ \left(\frac{4}{5}\right)^x > 0 $$ for all real $$ x $$. Therefore, both terms in $$ g'(x) $$ are negative (positive times negative), so $$ g'(x) < 0 $$ for all $$ x $$. This means $$ g(x) $$ is strictly decreasing, and hence $$ f(x) $$ is also strictly decreasing for all real $$ x $$.

Since $$ f(x) $$ is strictly decreasing and continuous, it can cross zero at most once. We already found one solution at $$ x = 2 $$.

Now, check the limits to confirm the behavior:
As $$ x \to \infty $$, $$ \left(\frac{3}{5}\right)^x \to 0 $$ and $$ \left(\frac{4}{5}\right)^x \to 0 $$, so $$ f(x) \to 0 + 0 - 1 = -1 < 0 $$.
As $$ x \to -\infty $$, let $$ x = -t $$ where $$ t \to \infty $$, so $$ f(-t) = \left(\frac{3}{5}\right)^{-t} + \left(\frac{4}{5}\right)^{-t} - 1 = \left(\frac{5}{3}\right)^t + \left(\frac{5}{4}\right)^t - 1 \to \infty $$ as $$ t \to \infty $$.

Thus, $$ f(x) $$ starts from $$ \infty $$ as $$ x \to -\infty $$, decreases strictly, crosses zero at $$ x = 2 $$, and goes to $$ -1 $$ as $$ x \to \infty $$. Since it is strictly decreasing and continuous, it crosses zero exactly once.

Therefore, the equation $$ f(x) = 0 $$ has exactly one solution.

Hence, the correct answer is Option C.