How to find an expression for bound charge densities in griffiths

[MrMuscle]

Attempt:

 

[Delta2]

The minus sign is lost because the differentiation is with respect to the source coordinates $r'$. It is a well known identity that $$\nabla\frac{1}{|\vec{r}-\vec{r'}|}=-\nabla'\frac{1}{|\vec{r}-\vec{r'}|}$$

where in the above the second $\nabla$ (the one in the right) has a $'$ next to it which means just that the differentiation is with respect to $r'$ coordinates.

For the rest you got the wrong identity involving curl and cross product. You should have the identity involving divergence and dot product which is as follows
$$\nabla\cdot (f\vec{A})=f\nabla\cdot\vec{A}+\vec{A}\cdot\nabla f$$

this identity holds for differentiating with source coordinates as well that is with $\nabla'$ in the place of $\nabla$. Apply this Identity for $\vec{A}=\vec{P}$ and $f=\frac{1}{|\vec{r}-\vec{r'}|}$.