Newtonian potential expansion identity

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.