- User joshphysics has already correctly answered OP's 1st question.
2a) Concerning OP's 2nd question, one derives
$$\tag{A}i\hbar \delta(x-x^{\prime})~=~i\hbar\langle x | x^{\prime} \rangle ~=~\langle x | [\hat{x},\hat{p}] | x^{\prime} \rangle ~=~\langle x | \hat{x}\hat{p} | x' \rangle-\langle x | \hat{p} \hat{x} | x' \rangle~=~(x-x^{\prime})\langle x | \hat{p} | x^{\prime} \rangle ~\stackrel{(1)}{=}~-i\hbar(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}).$$
In other words,
$$\tag{B}\delta(x-x^{\prime})~=~-(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}),$$
which also follows by differentiating the identity
$$\tag{C} (x-x^{\prime})\delta(x-x^{\prime})~=~0 $$
wrt. $x$.
2b) Eq. (B) should not be divide on both sides wrt. $x-x^{\prime}$. The problem is essentially that the distribution $\frac{1}{x}\delta(x)$ is ill-defined. The argument goes roughly as follows. Recall that a distribution $u$ should be evaluated on smooth test functions $g$. For instance, if the distribution $u$ is the Dirac delta distribution, then
$$\tag{D}u[g] ~:= ~g(0) $$
by definition. One can in general not multiply two distribution, but one can multiply a smooth function $f$ with a distribution $u$. By definition
$$\tag{E}(f\cdot u)[g] ~:= ~u[fg]. $$
So if $u$ is the Dirac delta distribution, one gets
$$\tag{F}(f\cdot u)[g] ~:= ~f(0) g(0). $$
In OP's case $f(x)=\frac{1}{x}$, so $f(0)$ is ill-defined.