How would you prove this
$$(\nabla^2 + k^2)\left(\frac{e^{ikr}}{r}\right) = -4\pi \delta^3(\vec{r})$$
using the fact that $\nabla^2(\frac{1}{r}) = -4\pi \delta^3(\vec{r})$?
I'm getting tripped up with how to apply the $(\nabla^2+k^2)$ term and going around in circles.
HINT:
In distribution, we apply the product rule for the Laplacian, $\nabla^2(fg )=f\nabla^2 (g)+2\nabla (f)\cdot \nabla (g)+g\nabla^2(f)$, with $f=\frac1r$ and $g=e^{ikr}$ to find
$$\nabla^2\left(\frac{e^{ikr}}{r}\right)=\underbrace{\frac1r \nabla^2\left(e^{ikr}\right)+2\nabla \left(\frac1r\right)\cdot \nabla (e^{ikr})}_{\text{Now, expand this.}}+\underbrace{e^{ikr}\nabla \left(\frac1r\right)}_{=-4\pi \delta3(\vec r)}$$
Can you finish now?