What is the general term formula of this recurrence relation, if it exists? $$a_n=-\frac{(a_{n-1} + 1)^2}{a_{n-1}+2}\quad \quad a_1 = -\frac 12$$
I transformed to:
$$a_n = -a_{n-1} - \frac{1}{a_{n-1} + 2}$$
$$a_{n-1} = -a_{n-2} - \frac{1}{a_{n-2} + 2}$$
...
And then
$$a_n - a_{n-1} + a_{n-2} -\cdots  + \cdots - a_1$$
$$=$$
$$- a_{n-1} + a_{n-2} - ... + ... - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ... $$
Many terms cancel each other and:
$$a_n = - \frac{1}{a_{n-1} + 2} + \frac{1}{a_{n-2} + 2} - ... + ... + \frac{1}{a_{1} + 2} + a_1$$ (Here need to divide cases odd vs even)
And then I'm forever stuck. I can't find analytical form of the sum above
I try to get the value of the first few dozen terms to observe. I find nothing other than the terms are negative fractions that satisfy the recurrence relation.
Question is, is there a general term formula? How to find it?

This is an incomplete answer. 
The recurrence is given by:
$$
a_n=-\frac{(a_{n-1}+1)^2}{a_{n-1}+2}
$$
Here we can separate both numerator and denominator as independent series. Let's create two new series, for them $c_n$ and $d_n$ respectively:
$$
a_{n-1}=\frac{c_{n-1}}{d_{n-1}}
$$
So:
$$
a_n=-\frac{(\frac{c_{n-1}}{d_{n-1}}+1)^2}{\frac{c_{n-1}}{d_{n-1}}+2}\\
=\frac{-(c_{n-1}+d_{n-1})^2}{c_{n-1}+2d_{n-1}}
=\frac{-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2}{c_{n-1}+2d_{n-1}}\frac{1}{d_{n-1}}\\
$$
Hence we can define the new sequences as:
$$
c_n=-c_{n-1}^2-2c_{n-1}d_{n-1}-d_{n-1}^2\\
d_n=c_{n-1}d_{n-1}+2d_{n-1}^2
$$
Both are quadratic forms, very similar:
$$
c_n=
\begin{bmatrix}
    c_{n-1} & d_{n-1}
\end{bmatrix} 
\begin{bmatrix}
    p_{111} & p_{112}\\
    p_{121} & p_{122}
\end{bmatrix} 
\begin{bmatrix}
    c_{n-1}\\
    d_{n-1}
\end{bmatrix} \\
d_n=
\begin{bmatrix}
    c_{n-1} & d_{n-1}
\end{bmatrix} 
\begin{bmatrix}
    p_{211} & p_{212}\\
    p_{221} & p_{222}
\end{bmatrix} 
\begin{bmatrix}
    c_{n-1}\\
    d_{n-1}
\end{bmatrix} 
$$
with:
$$
\mathbf{p}_{1\cdot\cdot}=\begin{bmatrix}
    -1 & -1\\
    -1  & -1
\end{bmatrix} \\
\mathbf{p}_{2\cdot\cdot}=\begin{bmatrix}
    0 & 1/2\\
    1/2 & 2
\end{bmatrix} 
$$
Which can be summarized by renaming the numerator and denominator series as $\mathbf{x}_{1|n}$ and $\mathbf{x}_{2|n}$, and by using tensorial notation with implicit summation over repeated indexes, excepting the recurrence index:
$$
\mathbf{x}_{k_n}=\mathbf{x}_{i_{n-1}}\mathbf{p}_{k_ni_{n-1}j_{n-1}}\mathbf{x}_{j_{n-1}}
$$
This recurrence, though tensorial, can be expressed up to the first term $\mathbf{x}_{\cdot|1}=[-1,2]$, where the tensor $\mathbf{p}$ appears $2^{n-1}-1$ times:
$$
\mathbf{x}_{k_n}=\mathbf{x}_{i_{1}} \mathbf{p}_{i_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}} 
 \cdots 
\mathbf{p}_{k_ni_{n-1}j_{n-1}} 
\cdots  
\mathbf{x}_{i_{1}} \mathbf{p}_{j_{2}i_{1}j_{1}} \mathbf{x}_{j_{1}}
$$
This could be depicted as a repetitive product of matrices, with $P$ appearing $2^{n-1}-1$ times:
$$
x_n=x_1 P x_1\cdots P \cdots x_1 P x_1 
$$
Though this expression is not closed and surely very far from what the OP wanted, this is not a recurrence, because it depends only on the first value of the sequence, and is a repetitive aplication of multidimensional products over the initial value. 
