Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that
$$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$
where 
$$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$
Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$


My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this [link][1] which states that for $\tilde{g}=e^{2f}g$ we have
$$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$
where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$,
$$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$
Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that
$$-\Delta_{\mathbb{R}^n} \phi(x)  = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$
Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$


  [1]: https://math.stackexchange.com/questions/3305797/showing-that-the-conformal-laplacian-is-a-conformally-covariant-operator