Let $f:X\rightarrow Y$ be a smooth morphism between smooth schemes. Then there is an exact sequence
$$0\mapsto T\_{X/Y}\rightarrow T\_{X}\rightarrow f^{\*}T\_{Y}\mapsto 0$$
Now let us assume $f$ to be a birational morphism with smooth fibers, meaning that the exceptional locus of $f$ is the disjoint union of smooth divisors. 
**In this setting does the above exact sequence still exist?**



Tangent bundles and birational morphisms
Unless the morhism $f$ is etale, this exact sequence is not true in general. The problem is this: the differantial $df$ can be viewed as an injection $f^{\*}K\_{Y}\to \Omega^{1}\_{X}$. One defines $\Omega^{1}\_{X/Y}$ to be the cokernel of this injection, so that we have an exact sequence: $0\to f^{\*}K\_{Y}\to \Omega^{1}\_{X} \to \Omega^{1}\_{X/Y}\to 0$. So far is true, but when you dualize, you do not get the exact sequence $0\to T\_{X/Y}\to T\_{X}\to f^{\*}T\_{Y}\to 0$. Because $T\_{X}\to f^{\*}T\_{Y}$ is not surjective in general. There is an $Ext^{1}$ which comes in after dualizing which is the cokernel of this non-surjective map. So as Piotr pointed out, there will only exist an exact sequence $0\to T\_{X/Y}\to T\_{X}\to f^{\*}T\_{Y}$. 


