How do I solve integrations of type: $$ I=\int{ d\mathbf{r} f(\mathbf{r})g(\mathbf{r}')\mathbf{A}(\mathbf{r})\cdot \nabla \left( \delta(\mathbf{r}-\mathbf{r}')\right) } $$ here $f$ and $g$ are scalar functions, and $\mathbf{A}$ is a vector function.
I know that for simple integrals of form $\int{ d\mathbf{r} f(\mathbf{r})g(\mathbf{r}') ( \nabla \left( \delta(\mathbf{r}-\mathbf{r}')\right)}$, the answer is $-(\nabla f(\mathbf{r}))g(\mathbf{r})$.
it is similar to the result you have. you can consider the form : $$ \int_{\Omega} d\mathbf{r} \mathbf{F}.\nabla(\delta(\mathbf{r}-\mathbf{r}')) $$ which $F=f(\mathbf{r}) g(\mathbf{r}')\mathbf{A}(\mathbf{r})$. now from one of Divergence theorem corollaries : $$ \int_{\Omega} \mathbf{F}.\nabla g d\mathbf{r} = \int_{\partial \Omega} (g\mathbf{F}).\nu d\sigma - \int_{\Omega}g(\nabla.F)d\mathbf{r} $$ there are so many reasons for neglecting second term in most problems. so we get $$ \int_{\Omega} \mathbf{F}.\nabla (\delta(\mathbf{r}-\mathbf{r}')) d\mathbf{r} = -\int_{\Omega} (\delta(\mathbf{r}-\mathbf{r}')) \,(\nabla.F)d\mathbf{r} \\ =-\int_{\Omega} (\delta(\mathbf{r}-\mathbf{r}')) \,g(\mathbf{r}') \bigg(\nabla. \big(f(\mathbf{r}) \mathbf{A}(\mathbf{r})\big) \bigg)d\mathbf{r} \\ = -g(\mathbf{r}')\bigg(\nabla. \big(f(\mathbf{r}') \mathbf{A}(\mathbf{r}')\big) \bigg) $$ actually the one you mentioned also should be $-(\nabla f(\mathbf{r}))g(\mathbf{r}) |_{\mathbf{r}=\mathbf{r}'}$ .
if you're integrating over a volume $\Omega$ which $\partial \Omega \neq \varnothing$ , or functions $f$ and $g$ do not vanish on $\partial \Omega$ and also $\mathbf{r}' \in \partial \Omega$ ; then you should consider the boundry term too.