**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.

areconformallyflat) are not all flat. – Robert Bryant Jan 13 '12 at 0:33