A puzzle with derivative of delta-functions

Question

I will assume as a given the fact that in terms of complex variables $z,\bar{z}$ the following formula holds (normalization is not essential) $$\partial_{\bar{z}}\frac{1}{z}=\delta(z)$$ Then, by the usual rules $$\partial_{\bar{z}}\frac{1}{z^2}=-\partial_z\delta(z)$$

Now consider the following rational identity $$ \frac{1}{z^2(z-w)}=-\frac{1}{z^2w}+\frac{1}{(z-w)zw} $$ and act on it with $\partial_{\bar{z}}$. In the lhs we get $$ \partial_{\bar{z}}\left(\frac{1}{z^2(z-w)}\right)=-\frac{\partial_z\delta(z)}{z-w}+\frac{\delta(z-w)}{z^2}=\frac{\partial_z\delta(z)}{w}+\frac{\delta(z-w)}{w^2} $$ while the rhs gives $$ \partial_{\bar{z}}\left(-\frac{1}{z^2w}+\frac{1}{(z-w)zw}\right)=\frac{\partial_z\delta(z)}{w}+\frac{\delta(z-w)}{zw}+\frac{\delta(z)}{(z-w)w}=\quad\frac{\partial_z\delta(z)}{w}+\frac{\delta(z-w)}{w^2}-\frac{\delta(z)}{w^2} $$ So the first two terms agree with the lhs but the last term is unmatched. Usually this level of rigor when operating with delta-functions doesn't get me into trouble, but apparently not this time. So, what is wrong with the computation?

Answer 1

OK, I've found and error. It is the assumption that $$f(z)\delta'(z)=f(0)\delta'(z)$$ when I replaced $-\frac{\partial_{z}\delta(z)}{z-w}$ by $\frac{\partial_{z}\delta(z)}{w}$ . The correct formula is $$\delta'(z)f(z)=f(0)\delta'(z)-\delta(z)f'(0)$$. Using it, one can write $$-\frac{\partial_{z}\delta(z)}{z-w}=\frac{\partial_{z}\delta(z)}{w}-\frac{\delta(z)}{w^2}$$ so that everything is consistent.