Hessian of a function on Riemannian manifolds Let $(M,g,\nabla)$ be a Riemannian manifold with metric $g$ and Riemannian connection $\nabla$. The hessian of a function $f:M\to R$ is defined by:
$$H^f(X,Y)=g(\nabla_X\ \ \operatorname{grad} f,Y)$$where $X,Y\in \frak{X}$$ ( M )$.
My Question is 


*

*Given a positive function $f$ defined on $M$, is there a function $h:M\to R$ such that
$$H^h(X,Y)=\frac 1fH^f(X,Y)$$Edit: One can use any additional assumption on $f$ to guarantee the existence of $h$.

*Given 
$$H^h(X,Y)=H^f(X,Y)$$
is there a relation between both functions $f,h$.

 A: Regarding question (1): When $M$ is compact and $f$ is a positive $C^2$ function, such an $h$ exists if and only if $f$ is constant on each component of $M$.
Proof: Suppose $M$ is compact and $f\colon M\to \mathbb R$ is a positive $C^2$ function. If $f$ is constant on each component, clearly $h\equiv 0$ works. Conversely, suppose $h\colon M\to \mathbb R$ is a $C^2$ function such that 
$$H^h(X,Y)=\frac 1fH^f(X,Y).$$
Taking the trace of both sides, we conclude that $\Delta h = \Delta f/f$. Now compute
$$
\Delta ( h - \log f)= \frac{\Delta f}{f} - \left(\frac{\Delta f}{f} - \frac{|\text{grad}\, f|^2}{f^2}\right) = \frac{|\text{grad}\, f|^2}{f^2}.
$$
Integrating both sides over $M$, we conclude that $|\text{grad}\, f|^2/f^2\equiv 0$ (because the integral of a Laplacian on a compact manifold is zero). This implies that $f$ is constant on each component of $M$. $\square$
When $M$ is noncompact, it might be possible to find such an $h$. For example, if $M=\mathbb R$ (with the Euclidean metric) and $f(x)=e^x$, we can take $h(x) = \tfrac12 x^2$. At the moment, I can't think of any necessary or sufficient conditions on $M$ or $f$ in that case. 
A: Suppose that $M$ is compact. 
For (2): It follows, by taking trace, that $\Delta f = \Delta h$, i.e., $f-h$ is harmonic and, hence, constant. Conversely, if $f-h$ is constant, Hessians are equal, of course. 
For (1). Again, one obtains $\Delta h= \frac{1}{f}\Delta f$. WLOG, by subtracting a constant from $h$, we can assume that $h$ is negative on the entire manifold. Suppose that $f$ is a (nonconstant) eigenfunction of the Laplacian, $\Delta f= \lambda f$, $\lambda >0$, hence, 
$$
\Delta h= \lambda. 
$$ 
Using negativity of $h$ and repeating the same proof (integration by parts) as for the fact that compact manifolds do not carry nonconstant harmonic functions, one obtains that $\nabla h=0$, i.e., $h=$constant, implying that $f$ is constant as well, a contradiction. 
One should be able to play a similar game for noncompact manifolds and get a complete answer in this case too... 
Edit: I just realized that for the item 1 my answer has a problem: The eigenfunction $f$ on a compact manifold will never be positive as required by the OP. (This can be seen by restricting it to a closed geodesic in $M$.) I still think that there are counter-examples to the problem (with positive $f$) but I do not have any at this point.  
