A problem about stability of the origin in an ODE related to sign and parity of the order of the first non-zero $n$-th derivative

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I am trying to solve this exam question

Considering the scalar ODE

$\dot x = f(x), x \in \mathbb{R}$, with $f \in C^{\infty}(\mathbb{R})$ and $f(0)=0 $and all the derivatives of $f $up to a certain order $n-1$ with $n \in \mathbb{N}\setminus\{0\} $ are zero at $x=0$. The n-derivative of $f$ is such that $f^{(n)}(0) \neq 0$. Now $x=0$ is necessarilly an stable equilibrium point if

a) $n$ is odd and $f^{(n)}(0)<0$

b) $n$ is odd and $f^{(n)}(0)>0$

c) $n$ is even and $f^{(n)}(0)<0$

d) $n$ is even and $f^{(n)}(0)>0$

My try:

By Taylor expansion around $x=0$ we have that

$f(x)=\frac{1}{n!}f^{(n)}(0)x^n +o(x^n)$

so we have

$\dot x = \frac{1}{n!}f^{(n)}(0)x^n +o(x^n)$

If $n=1$, I get $x= exp(\frac{1}{n!}f^{(n)}(0)t +o(1)t +constant)$

If $n>1$, I get $\frac{1}{-n+1}x^{-n+1}= \frac{1}{n!}f^{(n)}(0)t +o(1)t+constant)$

I don't know if this last part if ok, but anyway I don't know where to go from here Can you please help me finish it?