I'll give you some hints that hopefully will allow you to do your homework yourself.
In order to check linearity, write down $3$ difference equations: one for the response $y_1[n]$ to an input signal $x_1[n]$, one for the response $y_2[n]$ to an input signal $x_2[n]$, and one for the response $y_3[n]$ to an input signal $ax_1[n]+bx_2[n]$. Multiply the first difference equation by $a$, the second one by $b$, and add them. If the resulting difference equation for the sequence $ay_1[n]+by_2[n]$ is the same as for $y_3[n]$, then you can conclude $y_3[n]=ay_1[n]+by_2[n]$, and, consequently, the system must be linear.
For checking time-invariance, write down the difference equation for the response $y_2[n]$ to the input $x[n-k]$. Then replace in the original difference equation the index $n$ by $n-k$ and check if the two difference equations are the same. If they are, the system is time-invariant.
To answer the question about BIBO stability, ask yourself what the output signal will be for $x[n]=\delta[n]$ and draw your conclusion.
Now that you've figured out the solution I would like to point out that you can also find an explicit (non-recursive) expression for $y[n]$:
$$\begin{align} y[0]&=x[0]\\
y[1]&=1\cdot x[0]+x[1]\\
y[2]&=2\cdot 1\cdot x[0]+2\cdot x[1]+x[2]\\
y[3]&=3\cdot 2\cdot 1\cdot x[0]+3\cdot 2\cdot x[1]+3\cdot x[2]+x[3]\\
&\vdots\\
y[n]&=\sum_{k=0}^n\frac{n!}{k!}x[k],\qquad n\ge 0\tag{1}
\end{align}$$
from which it is very straightforward to show that the system is linear and time-varying. Furthermore, from $(1)$ it should be clear that the system is not BIBO-stable.