This question is two-sided, and I'm not sure what you mean by "local diffeomorphism" so I'll treat both aspects. There is a local and a global version :

## Local version :

**Q1 :** Given a point $x$ in a Riemannian manifold $(M,g)$, can we find a constant curvature metric on a neighborhood of $x$ which is close to $g$ ?

First remark : since we want a local statement, zero curvature is as good as constant curvature here.

The answer to **Q1** is "yes", *without any restriction on the curvature*. This can be seen using normal coordinates centered at $x$ : in these coordinates, the metric at $x$ is euclidean and its distortion from being euclidean as one moves away from $x$ can be controlled (using that the curvature is bounded in a neighborhood of $x$, and that the injectivity radius at $x$ is positive).

*Edit :* The above statement is too complicated. As Anton's say in his comment to Agol's answer, use the exponential map ! The pull back of $g$ by the exponential map defines a riemannian metric on some neighborhood of the origin in $T_xM$ which is equal to $g_x$ at $x$, by continuity, this pullback stay close to the euclidean metric $g_x$ on $T_xM$ and this does the job.

We can refine the question then :

**Q1' :** What can be said about the size of the neighborhood we obtained ?

In this case we need to impose geometric restrictions on $(M,g)$. For instance, Cheeger and Anderson proved the following :

For $n\in\mathbb{N}$, $k\in\mathbb{R}$, $i>0$ and $\varepsilon>0$, one can find $\delta>0$ such that in a $n$-manifold of Ricci curvature greater than $k$ and injectivity radius greater than $i$, any ball of radius $\delta$ admits a flat metric which is $\varepsilon$-close to $g$ in $C^0$-norm.

See "$C^\alpha$-compactness for manifolds with Ricci curvature and injectivity radius bounded below."

The proof uses more elaborate machinery than just normal coordinates : harmonic coordinates are used. If you stick to normal coordinates you can obtain a similar result but with stronger geometric assumptions.

## Global version :

**Q2 :** Under which condition does a manifold with almost $k$ curvature admit a $C^0$-close metric of constant curvature $k$ ?

If you consider large spheres, they have almost zero cuvrvature but don't admit any flat metric, so you need to put some restrictions on the side of the manifolds.

An example of theorem you can get is the following :

For any $n\in\mathbb{N}$, $k\in\mathbb{R}$, $V>0$, $D>0$ and $\varepsilon>0$, there is a $\delta>0$ such that any $n$-manifold $(M,g)$ of diameter less than $D$, volume more than $V$, and sectional curvature between $k-\delta$ and $k+\delta$ admits a metric af constant sectional curvature $k$ which is $\varepsilon$-close to $g$ in $C^0$-norm.

The proof relies on Cheeger-Gromov compactness theorem for sequences of Riemannian manifolds. A (really) sketchy goes like that : we argue by contradiction, you take a sequence $\delta_i$ going to $0$, and you assume you can find a sequence of manifolds $(M_i,g_i)$ satisfying the hypothesis of the theorem with $\delta=\delta_i$ and not satisfying the conclusion of the theorem. Then up to a subsequence, the sequence has a limit which is a manifold of constant curvature $k$, by the very definition of Cheeger-Gromov convergence, this imply the for some $i$ large enough, $M_i$ admit a constant curvature $k$ metric $\varepsilon$-close to $g_i$, a contradiction.

The lower bound on the volume is necessary (at least in the $k=0$ case) the so called "infranilmanifolds" admit metrics of curvature as close as wanted to $0$ with diameter bounded above but no flat metric.

For the $k=1$ case, the bounds on the diameter is unnecessary because of Myers theorem.

For the $k=-1$ case, I don't know if the hypothesis can be weakened.