There is little to find about this topic, but so far I have gathered:
$$\int_{-\infty}^{+\infty} f(r-r')\,\nabla_{r'}\delta(r')\,\mathrm{d^3r'} = \nabla_rf(r)$$
which can be obtained by taking us of:
- partial integration
- the gradient theorem $\left(\int\nabla_{r'}\delta(r')\,\mathrm{d^3{r'}} = \delta(r')\right)$
- as well as $\nabla_{r'}f = -\nabla_r f$
- and the fact that $f$ vanishes for $r \to \pm\infty\\[12pt]$
Now to step it up could you theoretically simplify:
$$\int_{-\infty}^{+\infty} f(r-r')\,\Delta_{r'}\delta(r')\,\mathrm{d^3r'} = \quad ?$$
The Dirac Delta is not a function and the object $\int_{\mathbb{R}^3}f(\vec r-\vec r') \nabla_{\vec r'}\delta(\vec r')\,d^3r'$ is not an integral. Rather, the Dirac Delta is a distribution and the integral symbol represents a linear functional.
The derivative $d'$ of a distribution $d$ is defined such that for any function $\phi\in C_C^\infty$ the functional $\langle d',\phi\rangle =-\langle d,\phi\rangle$. Therefore, we have for $f\in C_C^\infty$
$$\begin{align} \langle \nabla_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=-\langle \delta,\nabla_{\vec r'} \tau_{\vec r}Nf\rangle\\\ &=-\left.\left(\nabla_{\vec r'}\tau_{\vec r}Nf(\vec r')\right)\right|_{\vec r'=0}\\\\ &=-\left.\left(\nabla_{\vec r'}f(\vec r-\vec r')\right)\right|_{\vec r'=0}\\\\ &=\nabla_{\vec r} f(\vec r) \end{align}$$
where $N$ is the negation operator that takes $f(\vec r')$ to $f(-\vec r')$ and $\tau_{\vec r}$ is the translation operator that takes $f(-\vec r')$ to $f(\vec r-\vec r')$.
Then, in an analogous fashion we can write
$$\begin{align} \langle \nabla^2_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=\langle \delta,\nabla^2_{\vec r'} \tau_{\vec r}Nf\rangle\\\ &=\left.\left(\nabla^2_{\vec r'}\tau_{\vec r}Nf(\vec r')\right)\right|_{\vec r'=0}\\\\ &=\left.\left(\nabla^2_{\vec r'}f(\vec r-\vec r')\right)\right|_{\vec r'=0}\\\\ &=\nabla^2_{\vec r} f(\vec r) \end{align}$$
If one is unfamilar with the notation of distributions, then one can conduct the analysis using heuristical, formal arithmetic as follows.
$$\begin{align}\require{cancel} \langle \nabla^2_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=\int_{\mathbb{R}^3}f(\vec r-\vec r')\nabla^2_{\vec r'}\delta(\vec r')\,d^3r'\\\\ &=\int_{\mathbb{R}^3}f(\vec r-\vec r')\nabla_{\vec r'}\cdot \nabla_{\vec r'}\delta(\vec r')\,d^3r'\\\\ &=\cancelto{0}{\int_{\mathbb{R}^3}\nabla_{\vec r'} \cdot \left(f(\vec r-\vec r') \nabla_{\vec r'}\delta(\vec r')\right)\,d^3r'}\\\\ &-\int_{\mathbb{R}^3}\nabla_{\vec r'} f(\vec r-\vec r)\cdot \nabla_{\vec r'}\delta(\vec r')\,d^3r'\\\\ &=-\cancelto{0}{\int_{\mathbb{R}^3}\nabla_{\vec r'}\cdot \left(\nabla_{\vec r'} f(\vec r-\vec r)\delta(\vec r')\right)\,d^3r'}\\\\ &+\int_{\mathbb{R}^3} \nabla^2_{\vec r'}f(\vec r-\vec r')\delta(\vec r')\,d^3r'\\\\ &=\nabla_{\vec r}^2f(\vec r) \end{align}$$
as expected!