How to show that $$[H(x + t) − H(x − t)]\,\delta'(t) = −2\delta(x)\delta(t)$$ where $\delta$ represents the Dirac delta function and $H$ represents the Heaviside function?
I have tried testing the convolution on both sides with a test function $\phi(x,t)$ but it doesn’t seem to work.
Define the 2D distributions $$ u[f]~:=~ \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [H(x+t)-H(x-t)],\tag{1}$$ and $$ v[f]~:=~ \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) \delta(t)~:=~\int_{\mathbb{R}^2} \! \mathrm{d}x ~ f(x,0),\tag{2}$$ for a test function $f$.
The product distribution $$w~=~uv~=0\tag{3}$$ is well-defined. In fact $w$ is the zero distribution. Proof of eq. (3): $$\begin{align} u[f]~:=~&vw[f]\cr ~=~& \int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [H(x+t)-H(x-t)]\delta(t)\cr ~=~&0. \end{align}\tag{4}$$
OP's title relation is a consequence of that the derivative $$0~=~\partial_tw~=~(\partial_tu) v + u(\partial_tv) \tag{5}$$ vanish as well, and that $$(\partial_tu) v~=~2\delta^2. \tag{6}$$ Proof of eq. (6): $$\begin{align}(\partial_tu) v[f]~=~&\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) \partial_t[H(x+t)-H(x-t)]\delta(t)\cr ~=~&\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t) [\delta(x+t)+\delta(x-t)]\delta(t)\cr ~=~&2f(0,0)~=~2\int_{\mathbb{R}^2} \! \mathrm{d}x \mathrm{d}t f(x,t)\delta^2(x,t)\cr ~=:~& 2\delta^2[f].\end{align} \tag{7}$$