Step 1: Know what makes a log valid.
We have $f(x)=\log_{x-1}(3x+1)$. A logarithm has two parts: the number inside (called the argument) and the small number written below (called the base). For the log to give a real answer, both parts must follow strict rules.
Step 2: Rule for the inside number.
The number inside a log must always be more than zero. So we need \[ 3x+1 > 0 \] This gives $x > -\tfrac{1}{3}$.
Step 3: First rule for the base.
The base of a log must also be more than zero. Here the base is $x-1$, so \[ x-1 > 0 \] which means $x > 1$.
Step 4: Second rule for the base.
The base of a log can never be equal to 1, because $\log_1$ has no meaning. So $x-1 \neq 1$, which gives $x \neq 2$.
Step 5: Combine the rules.
We need all three at once. The strongest of $x > -\tfrac{1}{3}$ and $x > 1$ is just $x > 1$. So far the set is $(1,\infty)$.
Step 6: Remove the bad point.
From $(1,\infty)$ we still must throw out $x=2$. Taking it out splits the set into two pieces. So the final domain is \[ \boxed{(1,2)\cup(2,\infty)} \]