$f(0)=0$ and $\lvert\,f^\prime (x)\rvert\leq K\lvert\,f(x)\rvert,$ imply that $f\equiv 0$. I'll start with the precise statement of the problem:

Suppose that $f:[0,b]\to\mathbb{R}$ is differentiable, $\,f(0)=0$, and
  that there exists a real number $K\geq 0$ such that 
  $\lvert\,f^\prime(x)\rvert\leq K\lvert\,f(x)\rvert,$
  for all $x\in [0,b]$. Prove that $f$ is identically zero.

The problem provides the following "hint":

First, show that there exists $\delta > 0$ such that $f$ is zero on $[0,\delta]$.

At first I thought the hint seemed a bit odd, as of course $\delta = b$ would answer the whole problem. But then I started thinking that the hint probably meant that, if we could show that $f$ is zero on even a small (right-sided?) neighbourhood of $0$, that that would be enough to show $f$ must remain zero. In particular, the inequality in the problem would force $\lvert f^\prime(\delta) \rvert \leq K\lvert f(\delta)\rvert=0\implies f^\prime(\delta)=0$. But I'm stuck with what to do next because this of course does not alone imply that $f(x)=0$ for $x>\delta$.
Proving the hint itself has also evaded me. The wording, to me, implies that there's some property ,other than whatever properties would prove the general problem, we can use to produce this specific $\delta$, because otherwise, the hint would be useless. My first thought was to use the fact that $f$ is continuous at $0$ (and $f(0)=0$): Given $\varepsilon>0$, there exists $\delta>0$ such that $0<x<\delta$ implies $\lvert\,f(x)\rvert<\varepsilon$. 
I find myself wanting to use the Mean Value Theorem at this point: there exists $c\in (0,\delta)$ such that $f^\prime(c)=\dfrac{f(\delta)}{\delta}$. We can use the given inequality to produce
$$
\frac{\lvert\, f^\prime(\delta)\rvert}{K\delta}
\leq\left|\frac{f(\delta)}{\delta}\right|
=\lvert\, f^\prime(c)\rvert\leq K\lvert\, f(c)\rvert <K\varepsilon.
$$
Now, we could of course use $\varepsilon/K$ in the definition of continuity so that the previous inequality chain ends with $\epsilon$ instead of $K\epsilon$, but that's not the real issue: getting 
$\lvert\, f(\delta)\lvert<\delta\varepsilon$ looks promising until I realize that $\delta$ depends on $\varepsilon$, so all this really says is that $\lim\limits_{\delta\to 0^+}f(\delta)=0$, which is quite unsurprising.
So I suppose I need help with the hint itself as well as how it implies the result of the problem. I will note that this problem comes from a problem set that is supposed to be on the Riemann integral, but I have no idea how integration would come up here. I think we could claim that either $f(x)\geq 0 \,\forall\, x$ or $f(x)\leq 0 \,\forall\, x$; otherwise, $f$ would have a zero at a point where its derivative is nonzero, contradicting the inequality in the problem. If we could show that $\int_0^b f(x)\,dx=0$, it would then follow that $f$ is zero on $[0,b]$, but I don't see what could be said about the integral from the assumptions we're given.
 A: A) A hint for the hint: Choose $\displaystyle \delta=\frac{1}{2K}$. Suppose that $f$ is not zero on $[0,\delta]$. Let $M={\rm Sup}\{|f(x)|, x\in [0,\delta]\}$. There exists $d\in [0,\delta]$, with $d>0$ such that
$|f(d)|=M>0$. There exists $c\in ]0,\delta[$ such that $f(d)=df^{\prime}(c)$. Now use your hypothesis. 
B) A hint to continue: Put $g(x)=f(x+\delta)$. 
A: If $f(x) > 0$ 
$$\left (\vert f(x) \vert e^{-Kx} \right)'= e^{-Kx} \left (f'(x)-Kf(x) \right ) \leq 0$$
similarly if $f(x) \leq 0$
$$\left (\vert f(x) \vert e^{-Kx} \right)'= e^{-Kx} \left (-f'(x)+Kf(x) \right ) \leq 0$$
which implies $\left (\vert f(x) \vert e^{-Kx} \right) \leq \left (\vert f(0) \vert e^{-K(0)} \right) = 0$ and hence $f(x) = 0$
(The above inequalities and approach are similar to Gronwall Lemma)
A: If $K=0$ then problem is solved so assume $K > 0 $ , there exists a partion $P=\{x_0,x_1,...,x_n\}$ for $[0,b]$ such that $$K(x_i-x_{i-1})<1$$Assume $||f||_\infty$ and $||f'||_\infty$ be supperimum of $|f|$ and $|f'|$ on $[x_0,x_1]$ it is obvious that $$||f'||_\infty \le K||f||_\infty$$  there exists $u$ such that $x_0<u \le x_1 $ and $$f(u)=||f||_\infty$$ now with the help of mean value theorem there exist $c$ such that $x_0<c \le u$ and $$||f||_\infty=|f(u)|=|f(u)-f(x_0)|=|u-x_0||f'(c)|<\frac{1}{K} K ||f||_\infty= ||f||_\infty$$ so $||f||_\infty=0 $ so f is zero on $[x_0,x_1]$ and similarly in every $[x_i,x_{i-1}]$.
A: Let $g(x)=\int_0^x\lvert \,f'(t)\rvert\,dt$. Then $g(x)\ge \lvert \,f(x)\rvert$, and therefore
$$
0\le g'(x)\le Kg(x),
$$
for all $x\in [0,b]$. Thus
$$
\mathrm{e}^{-Kx}\big(g'(x)-Kg(x)\big)\le 0,\quad\text{for all $x\in [0,b]$,}
$$
and hence
$$
\big(\mathrm{e}^{-Kx}g(x)\big)'\le 0,\quad\text{for all $x\in [0,b]$,}
$$
which implies that $\mathrm{e}^{-Kx}g(x)$ is decreasing, and therefore
$$
0\le \mathrm{e}^{-Kx}g(x)\le g(0)=0, \quad\text{for all $x\in [0,b]$.}
$$
This implies that $$0=g'(x)=\lvert\,f'(x)\rvert=0,$$ and finally that $f\equiv 0$.
Note. We have assumed that $f'$ is integrable. Without this assumption, we can still prove it: Let $h(x)=f^2(x)$. Then $h(0)=0$, $h(x)\ge 0,$ for all $x\in[0,b]$, and 
$$
h'(x)=2f(x)f'(x)\le 2\lvert\,f(x)\rvert\lvert\,f'(x)\rvert\le 2K \lvert\,f^2(x)\rvert=2K h(x).
$$
Hence
$$
\mathrm{e}^{-2Kx}\!\left(h'(x)-2Kh(x)\right)\le 0
\quad\text{or}\quad \left(\mathrm{e}^{-2Kx}h(x)\right)'\le 0,
$$
which means that $\mathrm{e}^{-2Kx}h(x)$ is decreasing. Thus
$$
0\le \mathrm{e}^{-2Kx}h(x)\le h(0)=0.
$$
Therefore $h(x)=0$, for all $x$, and consequently $f\equiv 0$.
A: The condition given implies that $-K \leq {d \over dx} \log |f(x)| \leq K$ wherever $f(x)$ is nonzero. But the set of $x$ for which $f(x)$ is nonzero is an open set, and therefore is a countable union of intervals if nonempty. If $I$ is one such interval and $x_0$ is an endpoint of $I$, then $\lim_{x \rightarrow x_0} \log |f(x)| = -\infty$ since $f(x_0) = 0$, which contradicts that $-K \leq {d \over dx} \log |f(x)| \leq K$.
