Newtonian potential expansion identity Preliminaries
Consider the Newtonian potential $$\frac{1}{|\vec x - \vec y|}$$ with $\vec{x}, \vec{y} \in \mathbb{R}^3$ and $|\vec{x}| = x > y = |\vec{y}|$. Its Taylor expansion is given by $$\frac{1}{|\vec x - \vec y|} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x},$$ which can also be written in terms of Legendre polynomials as $$\frac{1}{|\vec x - \vec y|} = \sum_{n = 0}^\infty P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}.$$ The key identity here is $$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}}$$ Using the key identity twice, we have $$\frac{(-1)^n}{n!} (\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = P_n (\hat{x} \cdot \hat{y}) \frac{y^n}{x^{n+1}} = \left(\frac{y}{x}\right)^{2n+1}  P_n (\hat{x} \cdot \hat{y}) \frac{x^n}{y^{n+1}} = \left(\frac{y}{x}\right)^{2n+1} \frac{(-1)^n}{n!} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}.$$
The Question
Is there a way to prove the identity $$\boxed{(\vec{y} \cdot\vec{\nabla}_\vec{x})^n \frac{1}{x} = \left(\frac{y}{x}\right)^{2n+1} (\vec{x} \cdot\vec{\nabla}_\vec{y})^n \frac{1}{y}}$$ directly, i.e., without resorting to Legendre polynomials?
 A: Based on Fabian's correction, here is the proof:
start with noticing that $|y\vec x+{\Lambda\over y} \vec y|^2=y^2 x^2 +\Lambda ^2 +2\Lambda \vec x \cdot \vec y= |x\vec y+{\Lambda\over x} \vec x|^2 $. Thus
$$\frac{1}{y|\vec{x}+\frac{\Lambda}{y^{2}}\vec{y}|}=\frac{1}{x|\vec{y}+\frac{\Lambda}{x^2}\vec{x}|}$$
We  need the translation operator $e^{\vec{a}\nabla_{b}}$ which has the property  $e^{\vec{a}\nabla_{b}}F(\vec{b})=F(\vec{b}+\vec{a})$ as long as $\nabla_b$ and $a$ commute ( this can be seen by noting that applying the expansion of this operator is actually the Taylor series). By applying this operator on ${1 \over y} $ we  can get $$x^{-1}e^{\Lambda(x^{-2}\vec{x}\cdot\vec{\nabla}_{\vec{y}})}\frac{1}{y}=\frac{1}{x|\vec{y}+\frac{\Lambda}{x^2}\vec{x}|}
  $$ and similarly
$$y^{-1}e^{(\Lambda y^{-2}\vec{y}\cdot\vec{\nabla}_{\vec{x}})}\frac{1}{x}=\frac{1}{y|\vec{x}+\frac{\Lambda}{y^{2}}\vec{y}|}$$ Thus
$$y^{-1}e^{(\Lambda y^{-2}\vec{y}\cdot\vec{\nabla}_{\vec{x}})}\frac{1}{x}=x^{-1}e^{\Lambda(x^{-2}\vec{x}\cdot\vec{\nabla}_{\vec{y}})}\frac{1}{y}$$
Identifying $O(\Lambda^n)$ terms yields the identity.
