It is well known that for any smooth bounded (connected) domain $\Omega\subset\mathbb R^d$ with $d\ge2$, we can define a Green's function $G:\Omega\times\Omega\to\mathbb R$ in $\Omega$ which is smooth on $\mathring\Omega\times\mathring\Omega\setminus\Delta$, such that 
$$ G\*\phi(x) = \int\_\Omega\! G(x,y)\phi(y)\,\mathrm dy$$ solves the Poisson equation $-\Delta(G\*\phi)=\phi$ in $\Omega$, and $G\*\phi=0$ on $\partial\Omega$. Distributionally, we have that $-\Delta\_x G(x,y)=\delta\_y(x)$, and that $G(x,y)=0$ for $x\in\partial\Omega$, $y\in\mathring\Omega$. We know that we may write
$$
G(x,y) = g(x-y) + f(x,y)
$$
where $$ g(x):=
\begin{cases}-\frac{1}{2\pi}\log|x| & d=2\\ 
\frac{\alpha\_d}{|x|^{d-2}} & d\ge3\end{cases}$$
for some smooth $f:\mathring\Omega\times\mathring\Omega\to\mathbb R$ obtained by solving the Laplace equation.


What I was wondering was whether we can assert that $\nabla\_x G(x,y)$ is nonvanishing on the boundary $x\in\partial\Omega$ for $y\in\mathring\Omega$.



Boundary behavior of Greens functions on smooth bounded (planar) domains
This follows from the so-called (Eberhard) Hopf Minimum Principle. If you have a positive (super-) harmonic function $u$ in a ball, and $u(z\_0)=0$ for some boundary point $z\_0$,
then the normal derivative at $z\_0$ is non-zero. This is a simple exercise:
If the ball ix $|x|<R$ then 
$$m(r)=\min\{u(x):|x|=r\}$$
is concave in a certain sense (concave with respect to $r^{2-m}$ when $m>2$
and with respect to $\log r$ when $m=2$, where $m$ is the dimension of the space; this is called the Hadamard 3-circles theorem),
 and $m(R)=0$, so $m'(R-0)<0.$


If your boundary is sufficiently smooth, you can touch every boundary point by a ball
which lies inside the domain.


(Similar arguments also show that if you can touch your boundary point by a ball from outside of the region then the normal derivative is finite at this point).


