# Deriving photon propagator

In Peskin & Schroeder's book on page 297 in deriving the photon propagator the authors say that

$$\left(-k^2g_{\mu\nu}+(1-\frac{1}{\xi})k_\mu k_\nu\right)D^{\nu\rho}_F(k)=i\delta^\rho_\mu \tag{9.57b}$$

With the solution given in the next line in equation (9.58) as

$$D^{\mu\nu}_F(k)=\frac{-i}{k^2+i\epsilon}\left(g^{\mu\nu}-(1-\xi) \frac{k^\mu k^\nu}{k^2}\right)\tag{9.58}$$

Which is the propagator. I can verify this equation by inserting $D^{\mu\nu}_F(k)$ into the first equation, but I have no idea how to actually solve $D^{\nu\rho}_F(k)$ from $(9.57b)$. If anyone can help, it would be much appreciated.

$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ with A and B two unknown functions of the scalar k^2. The two tensor after A and B are the only possible Lorentz invariant tensors . Simply plugin and calculate the unknown functions.
It's just tensor equation reads: $$A_{\mu\nu}D^{\nu\rho}=i\delta_\mu{}^{\rho}$$, where $$A_{\mu\nu}=-k^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)k_\mu k_\nu$$. What we need to do it to find its Inverse $$A^{\mu\nu}$$. Certainly you can find it by brute force with the methods given in linear algebra, but it is easier to guess the answer.