Time invariance and linearity of recursive system
The system is both linear and time-invariant. 

Theory says linearity means that for an input $$ax_1[n] + bx_2[n]$$ your output would be $$ay_1[n] + by_2[n]$$ where $y_1[n], y_2[n]$ are the outputs of $x_1[n], x_2[n]$, respectively.

In your case, an input $x_1[n]$ gives output $$y_1[n] = y_1[n-1] - y_1[n-2] - x_1[n]$$ same for an input $x_2[n]$, thus $$y_2[n] = y_2[n-1] - y_2[n-2] - x_2[n]$$ 

Multiplying the first equation by $a$ and the second by $b$ we get $$ay_1[n] = ay_1[n-1] - ay_1[n-2] - ax_1[n]$$ and $$by_2[n] = by_2[n-1] - by_2[n-2] - bx_2[n]$$

Adding together, we get 
$$\begin{align}
ay_1[n] + by_2[n] &amp;= ay_1[n-1] - ay_1[n-2] - ax_1[n] + by_2[n-1] - by_2[n-2] - bx_2[n] \\
ay_1[n] + by_2[n] &amp;= ay_1[n-1] + by_2[n-1] - ay_1[n-2] - by_2[n-2] - ax_1[n] - bx_2[n])
\end{align}$$
which is the given system with $x[n] = ax_1[n] + bx_2[n]$ and $y[n] = ay_1[n] + by_2[n]$. So it's linear.

For time-invariance, we first delay the input by $k$, thus we get $$y[n] = y[n-1] - y[n-2] - x[n-k]$$

Can you show that for a delayed output (with same delay $k$), the same result is obtained?