If x is 0, f(x) = 1/√9 = 1/3.
As x increases, the denominator gets smaller, so f(x) gets larger. The function continues to grow "without bound" as x approaches 3, and when x is very close to 3, the function is very large. This continues until x = 3, at which point the denominator is undefined.
When x > 3, the expression inside the square root is negative. I'm guessing that you're not allowing for complex numbers, so let's say that when x > 3, the function is also undefined.
Now as x decreases from 0, the behavior of the function is the same -- it grows very large until x = -3, at which point it becomes undefined.
So the range is all real numbers greater than or equal to 1/3.