Planning of the question:


Let $(M,g)$ be a Riemannian manifold and $TM$ be its tangent bundle


The isotropic almost complex structures $J\_{\delta , \sigma}$ were introduced by Aguilar on the tangent bundle of a Riemannian manifold $(M,g)$ in [this paper](http://link.springer.com/article/10.1007%2FBF02568316#page-1). These structures induce a class of Riemannian metrics $g\_{\delta ,\sigma}$ on $TM$ which are a generalization of the Sasaki metric ($\sigma, \delta :TM \to \mathbb{R}^+$ are real positive functions). 


I supposed that the base manifold is the Euclidean space $\mathbb{R}^n$ and the Riemannian manifold $(T\mathbb{R}^n,g\_{\delta ,0})$ is an Einstein manifold with the Ricci operator $Ric(A)=\rho A$ where $A$ is a tengent vector to the manifold $T\mathbb{R}^n$ and $\rho =constant$. Then, I concluded the following necessary conditions for $\delta = \frac{1}{\alpha}$:


‎\begin{align}‎
‎&\sum \_{i=1}^n\frac{\partial ^2\alpha}{\partial (y^i)^2 }=-2\rho‎ , ‎\\‎ 
‎&\frac{1-n}{2 \alpha}(\frac{\partial \alpha}{\partial y^j})^2-\frac{\partial ^2 \alpha}{\partial (y^j)^2}=2\rho‎ , ‎\hspace{2mm}j=1,...,n,\\‎
‎&\frac{1-n}{2 \alpha}\frac{\partial \alpha}{\partial y^j}\frac{\partial \alpha}{\partial y^i}-\frac{\partial ^2 \alpha}{\partial y^j\partial y^i}=0,\hspace{2mm}i,j=1,...,n \hspace{2mm}\textit{and} \hspace{2mm}i\neq j‎,\\
&\frac{\partial \alpha}{\partial x^i}=0,\qquad i=1,...,n,
‎\end{align}‎
where we suppose that $\alpha = \frac{1}{\delta}$ and $(x^1,...,x^n)$ and $(x^1,...,x^n,y^1,...,y^n)$ are the standard coordinate systems on $\mathbb{R}^n$ and $T\mathbb{R}^n$, respectively(the summation on the indices is considered only in the first equation).


Question: Is there any function $\alpha: T\mathbb{R}^n \to \mathbb{R}^+$ satisfying the mentioned system? The locally answers are acceptable, too.



An answer to this system of PDE's
Unless I am somehow misunderstanding the notation, we can forget about the x dependence due to your last condition. So let $\alpha$ be a function of y. Your third equation can be written as
$${\partial^2 \over\partial y\_i\partial y\_j}(\alpha^{(n+1)/2})=0.$$
So it follows that
$$\alpha^{(n+1)/2}=\sum\_i f\_i(y\_i).$$
Now your second condition says that
$${\partial ^2\over \partial y\_i^2}(\alpha^{(n+1)/2})=-{4\rho\over (n+1)}\alpha^{(1-n)/2}.$$
Since the left hand side depends only on $y\_i$, this is possible only if $\alpha$ is also a function of only $y\_i$. But if this applies for every $i$, then $\alpha$ must be constant.


