jmc

The discriminant of the quadratic is $3^2-4(3)=-3<0$, so the quadratic has no real roots and is always positive for real inputs. The function is undefined if $0\le x^2+3x+3<1$, which since the quadratic is always positive is equivalent to $x^2+3x+3<1$.

To find when $x^2+3x+3=1$, we switch to $x^2+3x+2=0$ and factor as $(x+1)(x+2)=0$, so $x=-1$ or $x=-2$. The new quadratic is negative between these points, so the quadratic $x^2+3x+3$ is less than $1$ between these points, which makes the function undefined. So the domain of $f(x)$ is $$x\in \boxed{(-\infty ,-2]\cup [-1,\infty )}.$$