Question: $$t\_0=-5, t\_1=\frac{11}{5} $$

$$t\_n=1-\frac{2}{t\_{n-1}} + \frac{-4}{t\_{n-2}\*t\_{n-1}} $$

I have never seen so uncommon formula for recurrence like this before. I have no idea how to solve it. Please, can anyone help me ?

Answer: 
The recurrence can be re-written as $t\_n t\_{n-1} t\_{n-2} = t\_{n-1}t\_{n-2} - 2t\_{n-2} -4$.

Let $x\_n = t\_0 t\_1 t\_2 \dotsb t\_n$. Then
$$\frac{x\_n}{x\_{n-3}} = \frac{x\_{n-1}}{x\_{n-3}} - 2\frac{x\_{n-2}}{x\_{n-3}} -4$$

Hence $x\_{n} = x\_{n-1} - 2x\_{n-2} - 4x\_{n-3}$ and $x\_0 = -5$, $x\_1 = -11$, $x\_2 = -5$. You can find $x\_n$ from this (one solution of the characteristic equation is $r=-1$), and then $t\_n = \frac{x\_{n}}{x\_{n-1}}$.
I propose a solution utilizing the very interesting auxiliary sequence called $x\_n$ introduced by @Catalin Zara, verifying

$$x\_{n} = x\_{n-1} - 2x\_{n-2} - 4x\_{n-3}$$

The characteristic equation $r^3 - r^2 + 2r + 4 = 0$ of this recurrence relationship has the following roots $r=-1$, $r=1\pm i \sqrt{3}$.

Therefore, its general term can be expressed, using calculations that we do not reproduce here, in the following way 

$$x\_n=-\dfrac{1}{7}\left( A(-1)^n + B 2^n \cos(n\pi/3)+C 2^n \sin(n\pi/3)\right) \ \ \text{with} \ \ \ A=3, B=32, C=16 \sqrt{3}$$

(taking initial conditions into account). Therefore:

$$t\_n=\dfrac{x\_n}{x\_{n-1}}=\dfrac{A(-1)^n + B 2^n \cos(n\pi/3)+C 2^n \sin(n\pi/3)}{A(-1)^{n-1} + B 2^{n-1} \cos((n-1)\pi/3)+C 2^{n-1} \sin((n-1)\pi/3)}$$

Division of the numerator and the denominator by $2^{n-1}$ prove that this sequence is arbitrarily close to the following sequence:

$$u\_n:=2\dfrac{B \cos(n\pi/3) + C \sin(n\pi/3)}{B \cos((n-1)\pi/3)+C \sin((n-1)\pi/3)}$$

which has a cycle of length three: $2.5, 0.4,-8$. 

For example, $u\_{30}=\dfrac{2 B}{\dfrac{B}{2}-C\dfrac{\sqrt{3}}{2}}=-8$

Remark: this period-3 cyclic behavior persists when changing initial conditions.