**Assume given a smooth manifold $(\mathbb{R}^n, g)$, where the metric is a scaled identity $g = e^{2f}I$.**
**Is there a way to know if this is *always* a non-positive (sectional) curvature manifold?** 


Note this is a parametrized manifold that is locally conformally flat. 
Following *Einstein Manifolds* [Arthur L. Besse], the Ricci tensor (in coordinates), can be shown to be:


$R = -(n-2)(H\_f - \nabla f \cdot \nabla f^T ) - \frac{n-2}{n}(\Delta f + \|\nabla f\|^2)I\_{n\times n}$


where $H\_f$ is the Hessian of $f$. 


Then $(\mathbb{R}^n, g)$ is of non-positive (sectional) curvature if $R$ is negative semi-definite. 



Conformally-flat
I'm not quite sure what you mean by *always* non-positively curved. If you are asking if this metric is non-positively curved for any $f$ then this is false. If you are asking for conditions on $f$ ensuring that the resulting metric is non-positively curved then there is a general formula:


Let $(M,g)$ be a Riemannian manifold and let $\tilde g=e^{2f}g$ be a new metric on $M$. Let $p\in M$ and let $u,v$ be orthonormal with respect to $g$ vectors in $T\_pM$ and $\sigma$ the 2-plane spanned by them. 


Then $e^{2f}\tilde{K}\_\sigma =K\_\sigma-[Hess\_f(u,u)+Hess\_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2]$.


This formula is in Besse btw (Theorem 1.159) but it's written slightly differently there.


In the special case you are interested in $g$ is the canonical metric on $\mathbb R^n$ and hence $\tilde K$ is nonpositive iff $H\_f(u,u)+H\_f(v,v)+|\nabla f|^2-\langle \nabla f, u \rangle^2-\langle \nabla f, v \rangle^2\ge 0$ for any $p$ and any orthonormal $u$ and $v$ in $T\_pM$. Note that for example it's always true if $f$ is convex.


