For those who are interested, one can solve this using $\liminf$, $\limsup$ (So this does not quite meet the OP's requirement). The idea is described in another answer linked in the comment.
First of all one can show that $a_n$ is bounded. Let
$$\ell = \liminf_{n\to \infty} a_n, \ \ L = \limsup_{n\to \infty} a_n.$$
By the definition of $\ell$, one can find a subsequence $\{a_{n_k}\}$ so that $a_{n_k} \to \ell$. By picking a subsequence, we also assume that
$$a_{n_k - 1} \to \ell_1,\ \ a_{n_k -2} \to \ell_2\ \ \ \text{as }k\to \infty.$$
Then taking $k\to \infty$ of
$$a_{n_k} = \frac{1}{a_{n_k-1}} + \frac{1}{a_{n_k-2}},$$
we have
$$\ell = \frac{1}{\ell_1 } + \frac{1}{\ell_2} \ge \frac{1}{L} + \frac{1}{L} = \frac{2}{L} \Rightarrow \ell L \ge 2$$
Similarly by taking a subsequence going to $L$, we have $\ell L \le 2$. Thus $\ell L = 2$.
Now we show that $\ell =L =\sqrt 2$. Then this implies $a_n \to \sqrt 2$ as $n\to \infty$.
Similarly, let $a_{n_k}$ converges to $L$ and $a_{n_k-1} \to \ell_1$, $a_{n_k -2} \to \ell_2$ and $a_{n_k - 3} \to \ell_3$. Then we have
$$\frac {2}{\ell} = L = \frac{1}{\ell_1} + \frac{1}{\ell_2}$$
and
$$\ell_1 = \frac{1}{\ell_2} + \frac{1}{\ell_3}.$$
The first equality actually forces $\ell_1 = \ell_2 = \ell$. Put this into the second equality gives
$$\frac{2}{L} = \ell = \frac{1}{\ell } + \frac{1}{\ell_3}.$$
This again forces $\ell = \ell_3 = L$. Thus $\ell = L = \sqrt 2$.
You have for all $k \geq 1$
$$a_{k+1}-a_k = 2k$$
so if you sum these equalities for $k=1, ..., n-1$, you get
$$\sum_{k=1}^{n-1} (a_{k+1}-a_k) = \sum_{k=1}^{n-1} 2k$$
i.e.
$$a_n-a_1 = n(n-1)$$
i.e. finally because $a_1=-1$,
$$a_n=n^2-n-1$$
Best Answer
You need to investigate first whether $a_{n}$ is convergent or not. One way is by finding out whether it is bounded below/above and monotonically decreasing/increasing respectively (for $n>M$, for some $M$).
If you do have concluded that $a_{n}$ do converge to a value, say $a$. Then you can find $a$ by solving
$$ a = 3\frac{1+a}{3+a}$$
This is because in the long run (for large values of $n$), each sequence value will be 'the same' and equal to the limit.
Hope this helps.