I've stumbled across an older post here trying to solve the same problem the asker of the post had. The solution that was provided stated that for a harmonic function $u$ on $\mathbb{R}^n$ we have that $\nabla u (r\cdot0) = r\nabla u (0)$ for any $r > 0$. I'm not quite sure that this is true. Here's a counter example that I found:
$u((x,y)) = e^x \sin(y)$ is harmonic
$\nabla u((x,y)) = \begin{pmatrix} e^x \sin(y) \\ e^x \cos(y) \end{pmatrix} $
$\nabla u((r\cdot 0,r\cdot 0)) = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \neq r\begin{pmatrix} 0 \\ 1 \end{pmatrix} = r\nabla u((0,0))$
Am I missing something or is the claim indeed wrong?
You probably got the defintion of $u_r$ wrong. $u_r(x,y)=e^{rx}\sin (ry)$ and $\nabla u_r(0,0)=(0,r)$ and not $(0,1)$