(1) Proof: Let $x=y=0$, then $f(0) = f(0) + f(0)$, which gives us $f(0) = 0$.
Since the domain of function $f(x)$ is symmetric about the origin,
let $y=-x$, thus $f(0) = f(x) + f(-x)$.
With $f(0) = 0$, we get $f(-x) = -f(x)$.
Hence, the function $f(x)$ is an odd function.
(2) The function $f(x)$ is a monotonically increasing function on $\mathbb{R}$.
As given, $f(x) > 0$ when $x > 0$. Let's take $x_1 0$ implies $f(x_2 - x_1) > 0$,
which means $f(x_2) + f(-x_1) > 0$. Thus, $f(x_2) - f(x_1) > 0$.
Therefore, $f(x)$ is a monotonically increasing function on $\mathbb{R}$.
(3) Given $f(k \cdot 3^x) + f(3^x - 9^x - 2) < 0$,
we have $f(k \cdot 3^x) < -f(3^x - 9^x - 2)$, which implies $f(k \cdot 3^x) < f(-3^x + 9^x + 2)$.
Since $f(x)$ is a monotonically increasing function on $\mathbb{R}$, it follows that $k \cdot 3^x < -3^x + 9^x + 2$.
Therefore, we need $k < -1 + 3^1 + \frac{2}{3^x}$.
The inequality must hold for all $x \in \mathbb{R}$.
Hence, we need $k$ to be less than the minimum value of $-1 + 3^x + \frac{2}{3^x}$.
Then we note that $3^x + \frac{2}{3^x} \ge 2\sqrt{2}$.
Thus, $-1 + 3^x + \frac{2}{3^x} \ge -1 + 2\sqrt{2}$.
Therefore, $k < -1 + 2\sqrt{2}$,
which leads us to the conclusion that the range of $k$ is $\boxed{(-\infty, 2\sqrt{2}-1)}$.