Is this problem from Griffiths' "Classical electromagnetism" correct? I am talking about pbl 2.46 pag 107.
The potential $$\frac{A}{r} \, \exp{(-\lambda r)}$$ is given I have to find the charge density and the electric field.
The book says the solution for the charge density is: $$\left[\delta^3(r)-\dfrac{\lambda^2}{r} \, \exp{(-\lambda r)}\right]\, \epsilon_0\, A$$
The first part comes from $\nabla^2 \, \dfrac{1}{r}$ , the second comes from $\nabla^2\,\exp{(-\lambda r)}$. 
I think this part is missing: $2\,\nabla\, \dfrac{1}{r} \, \cdot\, \nabla\,\exp{(-\lambda r)} $. 
Or am I misunderstanding something?
 A: This is but the screened Poisson equation. Since the functions are radial, only the radial part of the Laplacian survives, 
$$
\nabla^2 (e^{-\lambda r}/r) = \frac{1}{r^2}\partial_r ( r^2 \partial_r (e^{-\lambda r}/r))= \frac{1}{r^2}\partial_r \left ( -\lambda r e^{-\lambda r} +e^{-\lambda r} r^2\partial_r \frac{1}{r}\right )\\
= \frac{1}{r^2}\partial_r \left(  r^2\partial_r \frac{1}{r}\right ) ~ e^{-\lambda r} -\lambda  e^{-\lambda r} \left(\frac{1}{r^2}+\partial_r \frac{1}{r} \right  ) + \lambda^2 \frac{ e^{-\lambda r}}{r}.
$$
Note the cross $O(\lambda)$ term vanishes.
Since
$$
\nabla^2 \frac{1}{r}=-4\pi \delta^{(3)} (\vec{r})= -\frac{\delta(r)}{r^2},
$$
one has 
$$
(\nabla^2 -\lambda^2) \left (\frac{e^{-\lambda r}}{r}\right)=-\frac{\delta(r)}{r^2}.
$$
So your text is just fine. You probably misunderstand, that, in fact, $\nabla^2 e^{-\lambda r}= (\lambda^2  -2\lambda/r )e^{-\lambda r}$, as  evident from above. This additional $O(\lambda)$ term here exactly cancels the cross term you wrote down. (Try evaluating $\nabla^2 r$ to see the point.)
