[This is a situation where things have been rewritten several times in response to an ongoing discussion. Let me try to reconstruct some of the temporal sequence here.]

**ROUND ONE**:

Ben's answer gives one way that your statement can fail: $Y$ can be singular and $f$ can be a birational morphism which does not induce an isomorphism of local rings at at least one of the singular points.

Here is something else that can go wrong: in positive characteristic, $f$ can be purely inseparable, e.g. the $p$-power Frobenius map.

**ROUND TWO**

I edited my response to point out that Ben's example has a nonreduced fiber over the singular point. I also said "I think" that mine does not, but this was pointed out by Kevin Buzzard to be false. [Or rather, the statement is false. I truly did think it was true for a little while.]

I also suggested that the following modification might be true:

Suppose $X$ and $Y$ are geometrically irreducible and $Y$ is **nonsingular** (together with all of the questioner's hypotheses, especially reducedness of the fibers!). Then if $f:X \rightarrow Y$ is a bijective morphism with reduced fibers, it is an isomorphism.

**ROUND THREE**

I typed up a counterexample over an imperfect ground field when the varieties are not geometrically integral (g.i. = the base change to the algebraic closure is reduced and irreducible: the reduced business has to be taken more seriously when the ground field is imperfect, since taking an inseparable field extension can introduce nilpotent elements). But Kevin Buzzard posted a simpler counterexample, so I deleted my answer.

**ROUND FOUR**

Kevin's answer also includes a beautifully simple example to show that the question is false even over $\mathbb{C}$ without some nonsingularity hypotheses: use nodes instead of cusps and remove one of the preimages of the nodal point.

I still wonder if my attempted reparation of the statement above is correct.