With $\delta^\mu_\nu$ the Kronecker delta, all indices running from $0$ to $3$, $g_{\mu\nu}=\text{diag}(1,-1,-1,-1)$, and $\delta^\mu_\mu =g_{\mu\nu}g^{\mu\nu}=4$, I am unable to show that
$$ D_{\nu\lambda}\left[ -\big( k^2-m^2 \big)g^{\mu\nu}+k^\mu k^\nu \right]=\delta^\mu_\lambda $$
leads to
$$ D_{\nu\lambda}=\dfrac{-g_{\nu\lambda}+\big(k_\nu k_\lambda\big)/m^2}{k^2-m^2}~~. $$
My intuition is to set $\lambda\to\mu$ so that $\delta^\mu_\lambda=4$ giving
$$ D_{\nu\lambda}\left[ -\big( k^2-m^2 \big)g^{\lambda\nu}+k^\lambda k^\nu \right]=4~~, $$
but here I become lost. Maybe I should multiply both sides by $g_{\lambda\nu}$, and I should probably complete a square or something to get that $m^4$ term floating around, but I am not seeing it. Any tips? Thanks.
The solution comes from an ansatz $D_{\nu\lambda}=Ag_{\nu\lambda}+Bk_\nu k_\lambda$.