Intuitively it seems that this should be the case but when attempting to apply this operator to a function $f$, we find
\begin{align*} \left(\int_{-\infty}^{+\infty} \frac{\partial}{\partial y} δ (y-x)\text{d} y\right)f(x) &= \int_{-\infty}^{+\infty} \frac{\partial}{\partial y} \left(δ(y-x)f(x)\right)\text{d} y \\ &= \int_{-\infty}^{+\infty} \left(δ'(y-x)f(x)+δ (y-x)\frac{\partial f(x)}{\partial y}\right) \text{d} y\\ &= \int_{-\infty}^{+\infty}\left(-δ(y-x)\frac{\partial f(x)}{\partial y}+δ(y-x)\frac{\partial f(x)}{\partial y}\right) \text{d} y\\ &= 0 \neq \frac{\partial f(x)}{\partial x} \end{align*}
I think even integration by parts interpretation works if you realize $\int \delta'(y-x) f(x) = \frac{\partial f(y)}{\partial y}$ and $\frac{\partial f(x)}{\partial y} = 0$. So now it works.
$$\int \delta'(y-x) f(x) = \int \lim_h \frac{(\delta(y+h-x)-\delta(y-x))}{h} f(x) $$ $$= \lim_h \frac{(\int\delta(y+h-x)f(x)-\int \delta(y-x)f(x))}{h} = \lim_h \frac{(f(y+h)-f(y))}{h} =\frac{\partial f(y)}{\partial y}$$