Let X be a compact kahler manifold, $M\subseteq X$ be a complex submanifold, is there a biholomorphism between a neighborhood of the zero section of the normal bundle $NM$ of $M$ and a neighborhood $U(M)\subseteq X$?
The existence of such a biholomorphism is rather rare. For example, as a simple exercise you can check that already for a conic in $\mathbb CP^2$ such a biholomorphism does not exist. At the same time, in the case $M$ is "exceptional" in $X$, for example, $X$ is obtained by a blow up from $X'$ at a point $x'\in X'$ and $M$ is the exceptional divisor, the situation is the often the one you want (I don't specify purposefully here what "exceptional" means). Solution to the exercise. Everything follows from the fact that each holomorphic map from $\mathbb CP^1$ to itself has at least one fixed point. Indeed if a neighbourhood of a conic $C$ in $\mathbb CP^2$ were biholomorphic to a neighbourhood of $O(4)$ bundle over $\mathbb CP^1$ then the restriction of $T \mathbb CP^2$ to $C$ would holomorphically decompose as a sum $N_C\oplus TC$ where $N_C$ is a line subbundle of the restriction. Now consider a line $L_x$ through each point $x$ of $C$ tangent to the direction $N_C(x)$. Finally consider $y(x)=L_x\cap C$ the second point of the intersection of $L(x)$ with $C$. We got the map $C\to C$ without fixed points, this is a contractioction. PS. I remember a theorem that states that the only smooth hypersurface in $\mathbb CP^n$ that has a neighbourhood isomorphic to a neighbourhood of a zero section in a line bundle is a linear $\mathbb CP^{n1}$. Thanks to jvp for the link below! 


The answer is "no" because in general there does not even exist a neighbourhood $U$ of your submanifold that holomorphically retracts to the submanifold.
Note that the hypotheses are very easy to satisfy: (1) holds for $K3$ surfaces for example and (2), (3) will follow from $X$ being sufficiently ample. Here is a freely downloadable version of Lárusson's article, from his homepage. 

