To control first derivative with the function itself: $f'(x)^2\leq Cf(x)$ near where $f(x_0)=f'(x_0)=f''(x_0)=0$. Let $f$ be a compactly supported nonnegative $C^2$ function. 
I want to show that there exists $C$, such that for all $x\in \mathbb R$, we have $f'(x)^2\leq  C f(x) $ by showing that for every point $x\in \mathbb R$, we can find a neighborhood $U$, that on $U$ we can find such a $C$.
However I have a trouble to find such a $U$ and $C$ when $f, f', f''$ all vanish at x. Any advice would be appreciated, thanks.
 A: For such $f$ there holds an estimate $f'(x)^2\le 2Mf(x)$ for all $x\in \mathbb R$, where $M=\max|f''|$. See lemma 5.6 in Yu. V. Egorov, Linear Differential Equations of Principal Type.
The idea is to consider $f$ on an interval $(x_1,x_2)$ where $x_1$ and $x_2$ are consecutive zeros of $f$. 
If the maximum of $f'^2/f$ on $[x_1,x_2]$ is attained at some  $c\in(x_1,x_2)$, then
$$2f''(c)f(c)-f'^2(c)=0$$ and $f'^2(x)\le 2f(x)|f''(c)|\,$ on $[x_1,x_2]$.
For endpoints L'Hopital's rule is used:
$$
\lim_{x\to x_1}\frac{f'(x)^2}{f(x)}=\lim_{x\to x_1}\frac{2f''(x)f'(x)}{f'(x)}=2f''(x_1).
$$
A: There is an $M>0$ with $$|f''(x)|\leq M\qquad\forall\>x\in{\mathbb R}\ .\tag{1}$$ 
Claim: $\quad f'^2(x)\leq 2M \>f(x)\quad\forall x\in{\mathbb R}$.
Proof. $\ $ When $f(x)=0$ at some point $x\in{\mathbb R}$ then $f'(x)=0$ as well, since otherwise $f$ would assume negative values in some points.
Therefore it is enough to show that $f(0)=:y_0>0$ and $f'(0)=-m_0$ with
$$m_0^2>2My_0\tag{2}$$
is impossible. Assume to the contrary that these conditions hold. From $(1)$ we then get
$$f'(x)\leq-m_0+Mx\qquad(x\geq0)\tag{3}$$
and therefore
$$f(x)=y_0+\int_0^x f'(t)\>dt\leq y_0+\int_0^x(-m_0+Mt)\>dt=y_0-m_0x+{1\over2}Mx^2\qquad(x\geq0)\ .$$
Putting $x:={m_0\over M}$ here it follows from $(2)$  that
$$f\left(m_0\over M\right)\leq y_0-{m_0^2\over 2M}<0\ .$$
From this we conclude that $f(\xi)=0$ for a $\xi$ with $0<\xi<{m_0\over M}$. At the same time, from $(3)$ it would  follow that $f'(\xi)<0$, which is impossible.
A: Caveat : there is a flaw in the proof below as explained in the comments.  
Your conjecture is false, as I show below.
My proof is conceptually straightforward but there are probably
more elegant and computationally simpler counterexamples.
If one removes the positivity condition on your function, a "canonical" counterexample
can be proved to exist by mere abstract algebra without any computation.
But we are lucky : the algebraic counterexample happens to be positive also
(hence the names "lucky lemma" below).
The basic idea is rather simple : we are going to construct a
$C^2$ function supported on $[0,2]$, such that 
$f(\frac{1}{n})=\frac{1}{n^3},f'(\frac{1}{n})=\frac{1}{n},
f''(\frac{1}{n})=0$ for any $n\geq 1$. 
Denote by ${\mathbb R}_5[X]$ the space of polynomials of 
degree $\leq 5$ in $X$.
Lemma 1. Let $a < b$ be two real numbers. Given a uple
$v\in{\mathbb R}^6$, there is a unique polynomial $P\in {\mathbb R}_5[X]$ 
such that $(P(a),P'(a),P''(a),P(b),P'(b),P''(b))=v$. We denote this
polynomial by $L(a,b,v)$.
Proof of lemma 1. We must show that
The linear map $\phi : {\mathbb R}_5[X] \to {\mathbb R}^6$,
$P\mapsto (P(a),P'(a),P''(a),P(b),P'(b),P''(b))$ is an isomorphism. Since the dimensions coincide, 
it suffices to show
 that the kernel of $\phi$ is trivial. But any $P\in {\sf Ker}(\phi)$
 must be a multiple of $(X-a)^3(X-b)^3$, and the conclusion follows.
Lemma 2 (Easy lucky lemma) The polynomial
$R=L(1,2,(1,1,0,0,0,0))$ is positive on $[1,2)$.
Proof of lemma 2. This polynomial is $R(x)=(2-x)^3\left((3x-\frac{7}{3})^2+\frac{5}{9}\right)$
Lemma 3 (Hard lucky lemma) The polynomial
$Q_n=L(\frac{1}{n+1},\frac{1}{n},(\frac{1}{(n+1)^3},\frac{1}{n+1},0,\frac{1}{n^3},\frac{1}{n},0))$ 
is positive on $[\frac{1}{n+1},\frac{1}{n}]$. 
Proof of lemma 3. We can write 
$$Q_n(x)=\frac{1-nx}{(n+1)^2}+\big(x-\frac{1}{n+1}\big)\bigg(\frac{2n+1}{n+1}(1-nx)+
\big(x-\frac{1}{n+1}\big)T_n(x)\bigg)$$ where $T_n$ is a polynomial of degree $3$ in $x$ ,
and $T_n\geq 0$ on $[\frac{1}{n+1},\frac{1}{n}]$ (I'll add more details on this last 
point later).
Finally, we can construct $f$ as follows :
$$
f(x)=\left\lbrace\begin{array}{lcl}
0, &\text{if} & x\leq 0, \\
Q_n(x), &\text{if} & x\in[\frac{1}{n+1},\frac{1}{n}],n\geq 1, \\
R(x), &\text{if} & x\in[1,2], \\
0, &\text{if} & x\geq 2. \\
\end{array}\right.
$$
