This is from a book by Srednicki:
\begin{eqnarray} [\Pi(x),H] &=& \frac{1}{2}\int{\rm d}^3y~[\Pi(x),\nabla^i\varphi(y)\nabla^i\varphi(y) + m^2\varphi^2(y)] \\ &=& \frac{1}{2}\int{\rm d}^3y~ \left( \nabla^i_y[\Pi(x),\varphi(y)]\nabla^i\varphi(y) + \nabla^i\varphi(y)\nabla^i_y[\Pi(x),\varphi(y)]\right. \\ &&~~~~~ + \left.m^2[\Pi(x),\varphi(y)]\varphi(y) + m^2\phi(y)[\Pi(x),\varphi(y)]\right) \\ &=& -i\int{\rm d}^3y~(\nabla^i_y\delta^3({\bf x}-{\bf y})\nabla^i\varphi(y)) + m^2\delta^3({\bf x}-{\bf y})\varphi(y)) \\ &=& -i\int{\rm d}^3y~ (-\delta^3({\bf x}-{\bf y})\nabla^2\varphi(y) + m^2\delta^3({\bf x}-{\bf y})\varphi(y))\\ &=& -i(-\nabla^2 +m^2)\varphi(x). \end{eqnarray}
The third equality is boggling me: it seems that he assumes that $\nabla^i(\delta(x-y))$ and $\nabla^i\phi(y)$ commute, but seems false to me. In fact, in the fourth equality he seems to use $\nabla^i\delta(x-y)=-\delta(x-y)\nabla^i$ (which seems to be in accordance to my notes on functional analysis), but if we apply this to the expression of the second equality, we get that the first term in the sum is $$ -\delta(x-y)(\nabla^i)^2\phi(y) $$ and the second one is $$ -\nabla^i\phi(y)\delta(x-y)\nabla^i $$ and these are not equal, as I suspected. So, what's going on here?
$\hat{\varphi}$ is an operator, whereas $\delta(x - y)$ is not, neither is its derivative. So
$$ \color{red}{\nabla \delta(x-y)}\color{blue}{\nabla\hat{\varphi}(y)} = \color{blue}{\nabla\hat{\varphi}(y)}\color{red}{\nabla \delta(x-y)} $$
In other words, they commute! So does $\hat{\phi}$ in the potential term
$$ \color{red}{\delta(x-y)}\color{blue}{\hat{\varphi}(y)} = \color{blue}{\hat{\varphi}(y)}\color{red}{ \delta(x-y)} $$