In (non-equilbrium) physics it is common to see the following transfromation done
$$v(x)\cdot \nabla_y \delta(x-y)\rightarrow \nabla_y\cdot\left\{ v(y)\delta(x-y)\right\}$$
even without the presence of integrals. What is the mathematical justification for this and why don't we have:
$$v(x)\cdot \nabla_y \delta(x-y)\rightarrow v(y)\cdot \nabla_y \delta(x-y)$$
In maths, distributions (say on $\mathbb{R}^n$) are viewed as linear functionals $T\colon \phi \mapsto T(\phi)\in \mathbb{C}$, defined on a suitable space of "test functions" $\phi\colon \mathbb{R}^n\rightarrow \mathbb{C}^{m}$. Given a suitably nice function $f$, there is an associated distribution $T_f\colon \phi \mapsto \int f \cdot \phi$. One commonly choses the test functions to be compactly supported, which allows for the following partial integration: $$ T_{\nabla f}(\phi)= \int \nabla f \cdot \phi = -\int f \mathrm{div}\phi = T_f(-\mathrm{div} \phi) $$ Since distributions are meant to generalize functions, one uses this as a definition for an aritrary distribution: $$ \nabla T(\phi):=T(-\mathrm{div}\phi) $$ In general it is not possible to multiply two distributions, but you can define the product of (nice) functions and a distribution by: $$ vT({\phi)=T(v\phi)} $$
Fix $x\in \mathbb{R}^n$. Then what you write as $\delta(x-y)$ can be interpreted as the functional $T \colon \phi \mapsto \phi(x)$. Then $$ \nabla(vT)(\phi)=vT(-\mathrm{div}\phi)=T(-v\mathrm{div}\phi)=v(x) \cdot (-\mathrm{div}\phi(x)) = v(x) \cdot T(-\mathrm{div}\phi)=v(x) \nabla T (\phi). $$ Note that $x$ is fixed, hence $v(x)$ is just a number. This works for an arbitrary test function $\phi$, hence we can say that $\nabla(v T)$ and $v(x) \nabla T$ coincide as distributions. In that sense your equation is mathematically justified.