Not sure if your equation is correct. I guess you meant
$$
y_t = \rho_1 y_{t-1} + \dots + \rho_p y_{t-p} + \epsilon_t
$$
In case of a AR(1) process you have to cast it into its MA($\infty$) (or 'covariance stationary') representation by reinserting the past observations $y_{t-j}$ where $j=1,\dots,\infty$:
$$
y_t = c + \rho_1 y_{t-1} + \epsilon_t .
$$
With $y_{t-1} = c + \rho_1 y_{t-2} + \epsilon_{t-1}$ it follows that
$$
y_t = c + \rho_1 (c + \rho_1 y_{t-2} + \epsilon_{t-1}) + \epsilon_t.
$$
If you do that infinitely you end up with
$$
y_t = \frac{c}{1-\rho} + \sum_{j=0}^{\infty} \psi_j \epsilon_{t-j} \quad \text{with} \quad \psi_j = \rho^j
$$
where we used that $\sum_{j=0}^{\infty}\rho^j c = \frac{c}{1-\rho}$ and the condition that $\rho < 1$.
The impsule-responses can now be calculated by
$$
\frac{\partial y_{t+j}}{\partial \epsilon_t} = \psi_j.
$$
You basically have to do the same for higher order AR(p) processes. There you can compute the $\psi_j$'s by 'comparing coefficients'. An important condition for an $AR(p)$ process to have a covariance stationary representation is that its roots all lie outside the unit circle (thats why we assumed that $\rho < 1$).
I hope this rudimentary info helps you...