First of all you can apply partial fraction expansion method to get the inverse DTFT to find the impulse resposne h[n] for this system. You have to assume casual ot anti-causal system.
$$H(e^{jw}) = \frac{3-e^{-jw}}{1 - 0.3e^{-jw} + 0.1e^{-j2w}} = \frac{3-e^{-jw}}{ (1 - \alpha e^{-jw})(1 - \alpha^* e^{-jw})}$$
$$H(e^{jw})= \frac{A}{1 - \alpha e^{-jw}} + \frac{B}{1 - \alpha^* e^{-jw}}$$
Where $\alpha = 0.1500 + j0.2784$ for this particular case. Hence assuming causality and using inverse DTFT look-up tables, deduce as:
$$\boxed{ h[n] = A \alpha^n u[n] + B (\alpha^*)^n u[n] }$$
Time domain recursion is also possible, requiring only two auxiliary values.
For example assuming causality one can use recursion from the LCCDE to obtain $y[n]$ for all $n \geq 0$ from the given LCCDE
$$y[n] - 0.3y[n-1] + 0.1 y[n-2] = 3x[n] - x[n -1]$$
by first rearranging it for $y[n]$ as
$$y[n] = 0.3y[n-1] - 0.1 y[n-2] + 3x[n] - x[n -1]$$
Then the recursion follows for $n \geq 0$ assuming that $x[n]=\delta[n]$
$$
\begin{align}
y[0] &= 0.3y[-1] - 0.1 y[-2] + 3x[0] - x[-1] \\
y[1] &= 0.3y[0] - 0.1 y[-1] + 3x[1] - x[0] \\
y[2] &= 0.3y[1] - 0.1 y[0] + 3x[2] - x[1] \\
y[n] &= 0.3y[n-1] - 0.1 y[n-2] \\
\end{align}
$$
Note that in computing $y[0]$ you need $y[-1]$ and $y[-2]$ as auxiliary (initial) conditions. Also note that $x[-1]=0$ and $x[0]=1$ and $x[n]=0$ for all remaining $n>0$.
A similar analysis fallows for the anti-causal case.
