Does $f(a)=1$ and $f'(x)=g\big(f(x)+x\big)f(x)$ imply that $f$ has no roots? (Proof verification) Let $f: I \to \Bbb R$ be differentiable, where $I$ is an open interval in $\Bbb R$. Let $a \in I$. Let $g: \Bbb R \to \Bbb R$ be continuous. If $f(a)=1$ and $f'(x) = g\big(f(x)+x\big) f(x)$, then prove that $f(x)=0$ does not have any solutions in $I$.
Is the following correct?
Let's say we have a $t$ that belongs in $I$ so that $f(t)=0$, then:
$$\lim_{x\to t}f'(x) = \lim_{x \to t} g\big(f(x)+x\big)f(x) \tag1$$
We know that $\displaystyle \lim_{x \to t} g\big(f(x)+x\big) = b$ where $b \in \Bbb R$, so $(1)$ becomes $\frac{\mathrm dy}{\mathrm dx} = b \ \mathrm dy$. Then, $b=\frac1{\mathrm dx}$, which is false because $b$ cannot tend to $\infty$. Can I cancel out $\mathrm dy$?
 A: For a rigorous proof lets recast our problem a bit. Assume $f:[-1,1]\to \Bbb R$ is differentiable, $f(0)=0$ but for $x>0$ you have $f(x)>0$. We want to show that there is a contradiction if there exists a continuous $g:\Bbb R\to \Bbb R$ so that $f'(x)=f(x)g(f(x)+x)$. If you want I can explain how solving this problem is the same as solving the above one.
Since $f(0)=0$ we can find a differentiable $h$ so that $f(x)=xh(x)$ (this is sometimes called the Morse Lemma, specifically $h(x)=\int_0^1 f'(tx)\,dt$). It follows that
$$f'(x)=h(x)+xh'(x)\overset!= f(x)g(f(x)+x)=xh(x)g(x(h(x)+1))$$
for $x>0$ divide by $xh(x)$ (because $f$ is non-zero here, so too must $h$) to get:
$$g(x h(x)+x)=\frac1x+\frac{h'(x)}{h(x)}\qquad\text{whenever }x>0$$
Now taking the limit $x\to0^+$ gives $g(0)$ on the left, so it should also do so on the right. First note that $\frac{h'(x)}{h(x)}=\partial_x\ln(h(x))$ and if the term on the right is to have a limit you must have that
$$\partial_x \ln(h(x)) = -\frac1x + c(x)$$
where $C(x)$ is some continuous function. So
$$\ln(h(x)) = \ln(1/x) + C(x)$$
where $C(x)$ is a continuous anti-derivative of $c$. It follows by taking exponentials (remember $h(x)$ is positive) that
$$h(x)=\frac{e^{C(x)}}x$$
for $x>0$. This is a contradiction to $h(x)$ being continuous at $0$, let alone differentiable.
