Time invariance of a System
Yes, this system is time-invariant assuming $x[n]\neq0 $ for all $n $.

The test for this is
$$x_{1}[n]=x[n-n_{0}] $$
$$y_{1}[n]=y[n-n_{0}] $$

So let's first express in terms of just $y[n]$.
$$ y[n]=\frac{x[n-3]-y[n-3]-y[n-1]}{x[n]} $$
Next, we go through the test.
$$y[n-n_{0}]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$
Substituting $y_{1}[n] $
$$y_{1}[n]=\frac{x[n-3-n_{0}]-y[n-3-n_{0}]-y[n-1-n_{0}]}{x[n-n_{0}]} $$
$$y_{1}[n]=\frac{x[n-3-n_{0}]-y_{1}[n-3]-y_{1}[n-1]}{x[n-n_{0}]} $$
And finally substituting $x_{1}[n] $
$$y_{1}[n]=\frac{x_{1}[n-3]-y_{1}[n-3]-y_{1}[n-1]}{x_{1}[n]} $$
Because the shifted sequence has the exact same relationship, it is said to be time-invariant. Most systems that don't alter $n$ or $t$ meet this. Be wary of anything multiplying or otherwise messing with the vector arguments beyond simple delays.