I think it is really natural to believe, after doing Riemannian geometry for a little time, that sectional curvature encodes the all local geometry of a Riemannian manifold. One of the first thing one learns is that having constant sectional curvature implies that you are locally isometric to either the sphere, the euclidian space or the hyperbolic space. 


The paper <http://www.jstor.org/stable/1970580> gives a satisfying answer to the question in dimension higher than 4 (Any diffeomorphism preserving the sectional curvature is an isometry, except in the constant case, which we already know to be locally determined by the curvature). (See also the MO question : [Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature](https://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature) for an interesting discussion on the question.)


Can anyone give a example of two smooth metric on a surface having the same Gaussian curvature function ? 


One could say : take two different hyperbolic compact surfaces of the same genus, their curvature functions are both -1, and yet are not globally isometric. But still those surfaces are locally isometric. 


So can one find two metric $g\_1$ and $g\_2$ on the disk inducing the same curvature function such that no diffeomorphism of the disk induces an isometry on a neighborhood $(U\_1,g\_1)$ of $0$ on $(U\_2=f(U\_1),g\_2)$. Which are essentially different is this sense. 



Examples of non isometric surfaces having the same curvature function
Yes, you can easily do this with surfaces of revolution: If you take a metric of the form $ds^2 = dr^2 + f(r)^2\,d\theta^2$, where $f$ is an odd function of $r$ satisfying $f'(0)=1$, then the Gauss curvature function $K(r)$ satisfies 
$$ f''(r) + K(r)\,f(r) = 0,$$ 
where $K(r)$ is an even function of $r$. 


Suppose now that $K\_1$ and $K\_2$ are any two even functions of $r$ that have a nondegenerate minimum of $0$ at $r=0$. Then there will be a unique smooth, increasing, odd function $\phi$ that satisfies $K\_1(r) = K\_2(\phi(r))$ (and $\phi(0)=0$). Now consider the mapping of the disk $|r|<\epsilon$ defined by $F(r,\theta) = (\phi(r),\theta)$, and let $f\_1(r)$ and $f\_2(r)$ be the odd functions of $r$ that satisfy $f'\_i(0)=1$ and
$$ f''\_i(r) + K\_i(r)\,f\_i(r) = 0.$$
Then $F$ aligns the curvatures of the two metrics $ds\_i^2 = dr^2 + f\_i(r)^2\,d\theta^2$, but generally isn't an isometry if $\phi$ is not the identity map. Moreover, it is not hard to show that there is *no* isometry between the two metrics. 


The two metrics $ds^2\_1$ and $F^\*(ds^2\_2)$ now satisfy your requirements.


**N.B.:** In addition to Kulkarni's paper, which the OP cites above, it is probably a good idea to look at Yau's 1974 Annals paper, "Curvature preserving diffeomorphisms" (<http://www.jstor.org/stable/1970843>), where he discusses why the result does not hold in dimension $3$. I might also add that you could find some useful information by looking at the answers to the MO question [Does the curvature determine the metric?](https://mathoverflow.net/questions/100281/does-the-curvature-determine-the-metric/100372#100372)


