Rearranging the given relation as $$3a_na_{n-1}+a_n+a_{n-1}=0 \implies a_{n-1}-a_n=3a_na_{n-1}$$ Because all terms are non zero, dividing both sides of the above equation by $a_na_{n-1}$ yields $$\frac{1}{a_n}-\frac{1}{a_{n-1}}=3$$ This implies $\{\frac{1}{a_n}\}$ is an arithmetic sequence whose first term is $\frac{1}{a_1}=1$ and common difference is $3$. It follows that $$\frac{1}{a_n}=1 + 3\times (n-1)=3n-2\implies \boxed{a_n = \frac{1}{3n-2}}$$