What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

$ce{H}$ is less electronegative than $ce{O}$, so you pretend the electrons in the $ce{H-O}$ bond "belong" to $pu{O}$; $ce{H}$ is $+1$. By similar reasoning, $ce{F}$ is $-1$.
You can work out the oxidation number of $ce{O}$, either by counting electrons, or by using the fact that adding up all the oxidation numbers of any chemical species has to give you the charge of that species, which in this case of course is $0$.

$ce{F}$ is the most electronegative element, so it is always -1 oxidation state, except in $ce{F2}$, where it is zero. So, in $ce{HOF}$, $ce{H}$ has +1, $ce{O}$ has zero and $ce{F}$ has -1 oxidation states. There is no way $ce{F}$ will ever have oxidation state +1. In $ce{OF2}$ (oxygen fluoride, and not fluorine oxide) $ce{O}$ has +2 oxidation state!

The oxidation state in my opinion will be zero.Because as flourine is the most electronegative element it will never show positive ooxidation state.It's maximum oxidation state is zero.

Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States.

When you apply the official pre-2016 definition (via the Internet Archive) from the IUPAC gold book, then you have to assign $text{OS}(ce{H}) = +1$ as it is not a compound with a metal, $text{OS}(ce{O}) = -2$ as it is no peroxide compound, and leaving the disturbing $text{OS}(ce{F})= colorred{+1}$.

Going with the on electronegativity based alternative description, you will still assign $text{OS}(ce{H}) = +1$ as it has the lowest electronegativity. Then you assign $text{OS}(ce{F})= -1$, because of the highest electronegativity. This leaves a oxidation state of $text{OS}(ce{O}) = 0$.

This assignment is matched by the 2016 definition, the summary version of which is: The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds. The definition refers to the use of Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail). Consequently the heteronuclear approximation yields:

The true charges are of course something completely different. Based on a DF-BP86/def2-SVP calculation I ran some population analyses with Multiwfn 3.4.1 (a newer version is available). The NBO charges are taken from a prior version of this answer, but I since have lost access to the program and the files used.

begin{array}{lrrr}hline text{Method} & ce{H} & ce{O} & ce{F} \hline text{Hirshfeld} & +0.18 & -0.10 & -0.09 \ text{ADCH} & +0.38 & -0.29 & -0.09 \ text{VDD} & +0.18 & -0.10 & -0.09 \ text{Mulliken} & +0.20 & -0.06 & -0.15 \ text{Löwdin} & +0.09 & +0.01 & -0.10 \ text{Becke} & +0.38 & -0.21 & -0.17 \ text{ADC Becke} & +0.38 & -0.29 & -0.09 \ text{CM5} & +0.35 & -0.26 & -0.08 \hline text{QTAIM} & +0.61 & -0.44 & -0.17 \hline text{NBO} & +0.45 & -0.32 & -0.12 \hline end{array}