Is there any function which grows 'slower' than its derivative?

expanded from David's comment

$f' > f$ means $f'/f > 1$ so $(\log f)' > 1$. Why not take $\log f > x$, say $\log f = 2x$, or $f = e^{2x}$.

Thus $f' > f > 0$ since $2e^{2x} > e^{2x} > 0$.

added: Is there a sub-exponential solution?
From $(\log f)'>1$ we get $$ \log(f(x))-\log(f(0)) > \int_0^x\;1\;dt = x $$ so $$ \frac{f(x)}{f(0)} > e^x $$ and thus $$ f(x) > C e^x $$ for some constant $C$ ... it is not sub-exponential.

There are multiple mistakes in your proof (i.e. dividing by $f(x)$ is not necessarily okay, since it is not necessarily positive). The most major is that you treat the variable in the limit as if it were not "bound". That is, if you have something like $$\lim_{h\rightarrow 0^+}1>0$$ which is true, you can't necessarily, say, multiply through by $h$ to get $$\lim_{h\rightarrow 0^+}h>0\cdot h$$ which is false. This is essentially what you do, and why your conclusion is wrong. You have to regard the $h$ as belonging to the limit - that is $\lim_{h\rightarrow 0^{+}}f(h)$ is a constant - it does not depend on $h$, because there is no notion of "$h$" outside of the limit.

For an example of a function that does not satisfy this, you can take $e^{\alpha x}$ for any $\alpha>1$. Another solution is $xe^x$. It's worth noting that since the solution to $f'=f$ is $x\mapsto e^x$ we can prove that any solution grows faster than exponentially.

Even more generally than what other answers have brought up (though this implicit in a step of GEdgar's proof), if you let $g$ be a continous function, and $G$ an arbitary primitive of $g$ (that is, G' = g), then if $f(x) = e^{G(x)}$, we have $f' = gf$. So the ratio of a function to it's derivative can grow at essentially arbitary rates.

Edit: as Marc van Leeuwen points out, an example gotten from this is that $f(x) = e^{x^2}$ satisfies the property that $f'(x)/f(x) = 2x$ so the derivative grows asymptotically faster than the function.