Solution #1 To determine the range, we suppose $ y=\frac{3x+1}{x+8} $ (where $ x\ne -8 $) and see if we can solve for $ x $: $$\begin{array}{r r@{~=~}l} & y & (3x+1)/(x+8) \\ \Leftrightarrow & y(x + 8) & 3x + 1 \\ \Leftrightarrow & yx + 8y & 3x + 1 \\ \Leftrightarrow & x(y - 3) & 1 - 8y.\end{array}$$This last equation gives a contradiction if $ y=3 $, since in this case it says that $ 0=-23 $. Therefore, it is impossible for $ g(x) $ to equal $ 3 $ for any value of $ x $. But for any value of $ y $ other than $ 3 $, the last equation can be solved to yield $ x = \frac{1-8y}{y-3} $, or, in other words, $ g\left(\frac{1-8y}{y-3}\right)=y $. Therefore, the range of $ g(x) $ is $ \mathbb{R}\setminus\{3\} = \boxed{(-\infty,3)\cup(3,\infty)} $. Solution #2 We can rewrite $ g(x) $ as follows: $$g(x) = \frac{3x+1}{x+8} = \frac{3x+24}{x+8}-\frac{23}{x+8} = 3 - \frac{23}{x+8}.$$Then we note that $ x+8 $ takes on all real values, so $ \frac{1}{x+8} $ takes on every value which is the reciprocal of some nonzero real number, i.e., $ \frac{1}{x+8} $ takes on all nonzero values. Accordingly, $ 3-\frac{23}{x+8} $ takes on all values not equal to $ 3 $. Therefore, the range of $ g(x) $ is $ \mathbb{R}\setminus\{3\} = \boxed{(-\infty,3)\cup(3,\infty)} $.