If $f$ is everywhere right-differentiable, then the two right side derived numbers $\Lambda_r,\lambda_r$ are equal and finite at each $x$. These two derived numbers are defined as respectively the lim sup and lim inf of $(f(x+h)-f(x))/h$ as $h \to 0^+.$ There are similarly defined left side derived numbers $\Lambda_l,\lambda_l,$ and the Denjoy-Young-Saks Theorem relates these for arbitrary functions. According to the theorem, except for a set of measure zero two associated (same side) derived numbers are either equal and finite or unequal with at least one infinite, while two opposed (opposite side) derived numbers are either finite and equal or infinite and unequal.
With our assumptions the two right side derived numbers are equal and finite everywhere.
Applying what the theorem says about opposed derived numbers, the left derived numbers must each equal the common value of the right side derived numbers. (We do not have either right side derived number infinite). So all four derived numbers are equal and $f$ is differentiable (except for a set of measure zero).
Source for the Theorem: Functional Analysis, Riesz and Sz-Nagy, pp17-19
There's a wiki page for the theorem, with different notation for the derived numbers, here
Added note: I wasn't careful about what the term "opposed" derived numbers means, since for this problem it made no difference. A lim sup on one side is considered "opposed" only to the lim inf on the other. The example of $f(x)=x$ for rational $x$ else $2x$ for irrational $x$ shows we need to consider "opposed" as just mentioned. At nonzero rational $x$ we have $\lambda_r=\Lambda_l=1$ while $\Lambda_r=+\infty$ and $\lambda_l=-\infty.$ The same kind of thing occurs at irrational $x$ with the finite coinciding value $2.$ At $x=0$ only we have the opposing pairs unequal, one is $1$ and the other is $2,$ so for this function $0$ is in the set of measure zero for which the set of possibilities for the derived numbers doesn't hold.
About weakening the condition: I think if we just assume the right derivative of $f$ exists almost everywhere, the above argument goes through and one can conclude $f$ is differentiable almost everywhere. One is only throwing out a set of measure zero to begin with, and then the above can be applied to the remaining part of the real line.
An interesting thing would be to suppose only that the right derivative of $f$ exists at each point of a set $A$ which is countable and dense in the reals. Then the above theorem doesn't apply, but it seems difficult (to me) to come up with an example wherein $f$ fails to be differentiable, at least on some dense subset $B$ of $A.$