lim x^2-9/x+3 = -6
x-->3

I think that you meant:
\lim_{x \to -3}\ \frac{x^2 - 9}{x+3} = -6

What happens if you factorise the denominator? :)

Please use proper grouping symbols. You are in calculus.

Are you sure there is not a typo.
x should perhaps approach -3 instead of 3?

Here's why I ask:

The numerator is a difference of two squares.

\frac{(x+3)(x-3)}{x+3}=x-3

\lim_{x\to -3}(x-3)=-6

If x approached 3, the limit would be 0.