It is well known that $\delta'(x)\phi(x)=-\delta(x)\phi'(x)$ in a distributional sense, for some well behaved test function $\phi(x)$. However, what about the case:
$$\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\int_{-\infty}^\infty dx \,\delta'(x+y+z)\phi(y)\psi(z)=~~???$$
Is this equal to zero? Or maybe it is equal to $\int_{-\infty}^\infty dz\int_{-\infty}^\infty dy\,(\phi'(y)\psi(z)+\phi(y)\psi'(z))$? Who knows?
In this case using the co-area formula define the plane $P_c$ by $x+y+z=c$ and its surface measure by $dS_c$, then
\begin{align} \int_{\mathbb{R}^3} \delta'(x+y+z) \phi(y) \psi(z) dx dy dz & = \frac{1}{\sqrt{3}} \int_{-\infty}^\infty \int_{P_c} \delta'(c) \phi(y) \psi(z) dS_c dc \\ & = -\frac{1}{\sqrt{3}} \frac{d}{dc} \left. \int_{P_c} \phi(y) \psi(z) dS_c \right |_{c=0} \end{align}
To finish the problem you should choose a parametrization of the plane, compute the integral with $c$ as a parameter and then differentiate. I think that since there is no other dependence on $x$ that you will actually get zero (by parametrizing the plane by $(y,z)$ taking $x=c-y-z$).