From Mathworld, for example, we have the following properties of the Dirac delta:
$x^n\delta^{(n)}(x)=(-1)^n\, n! \, \delta(x)$
$x^2 \, \delta'(x)=0$
So, if $f(x)$ is $C^\infty(R)$, is it correct to guess that, from its Taylor expansion:
$f(x) \, \delta''(x-a)=f(a)\delta''(x-a)-f'(a)\delta'(x-a)+f''(a)\delta(x-a)$?