Two identities involving Dirac deltas In A. Smilga (2001) Lectures in Quantum Chromodynamics (p. 10) the following two identities are presented:
$$\delta'(x-y)[f(y)-f(x)] = -\delta'(x-y)(f'(x)(x-y)) = \delta(x-y)f'(x)$$
but no proof is provided, how can we verify these identities.

My attempt is to define:
$$F(y):= \langle \delta'(x-y), f(y)-f(x)\rangle = \int_{-\infty}^{\infty} \delta'(x-y)[f(y)-f(x)]\text{d}x$$
assuming $f\in C^\infty_0(\mathbb{R})$, integration by parts gives us:
$$F(y):= -\langle \delta(x-y), -f'(x)\rangle = \int_{-\infty}^{\infty} \delta(x-y)f'(x)\text{d}x$$
This is, essentially, the second identity $\delta'(x-y)[f(y)-f(x)] = \delta(x-y)f'(x)$. But I am not able to devise how to proof the first identity. I wonder if the "formal" definition:
$$\delta(x-y)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ip(x-y)}\text{d}p$$
can be useful for both identities.
 A: $\newcommand{\R}{\mathbb{R}}\newcommand{\d}{\mathrm{d}}$

*

*The distribution $\delta'(x)$ is defined by
$$\delta'[f] := \int_\R\d{x}\ \delta'(x)f(x) = -f'(0),$$
which you can show (as you did) by integration by parts. Equivalently you can write
$$\delta'[f](y) := \int_\R\d{x}\ \delta'(x-y)f(x) = -\left.\frac{\d}{\d x}f(x)\right|_{x=y},$$

*By direct computation you can show that
\begin{align} F[f](y):= \int_\R\d{x}\ \delta'(x-y)\Big(f(y)-f(x)\Big) = -\frac{\d}{\d x}\Big(f(y)-f(x)\Big)\bigg|_{x=y} = f'(y)
\end{align}
is equal to
\begin{align} G[f](y)&:= \int_\R\d{x}\ \delta'(x-y)\Big(-f'(x)(x-y)\Big) = -\frac{\d}{\d x}\Big(-f'(x)(x-y)\Big)\bigg|_{x=y}=\\
&=\Big(f''(x)(x-y)+f'(x)\Big)\bigg|_{x=y}=f'(y).\end{align}
A: The cheapest, informal, seat-of-the-pants way to see this, avoiding excess formulas, is to recognize the power of differentiation,
$$
\delta(z) ~ z =0 \leadsto \delta ' (z) ~z + \delta (z) =0 \leadsto -\delta' (z) ~z = \delta(z),\\ 
\delta(z) ~ z^2 =0 \leadsto \delta ' (z) ~z^2 + \delta (z) 2z=0 =  \delta ' (z) ~z^2 ,
$$
etc for higher powers of z. Essentially, the odd generalized functions vanish at the origin.
The Taylor series of your expression around x then collapses,
$$\delta'(x-y)[f(y)-f(x)] = \delta'(x-y)~[ (y-x)f'(x) + (y-x)^2 f''(x) /2 + ...] \\ =
-\delta'(x-y)~ (x-y) ~f'(x)= \delta(x-y)f'(x),$$
by the above. This is what the author would expect  you to perform in your head, really.
