Doubt regarding one-one onto function

Suppose $f$ is continuous. Symmetric around origin means $f(-x)=-f(x)$ for all $x\in\mathbb{R}$. Taking limit as $x\to 0$ we get $f(0)=0$. We have $f(x_1)>0$ for some $x_1>0$. Take some $x_2>0$ and $x_2\neq x_1$. Then $f(x_2)$ cannot be $0$ because $f(0)=0$ and $f$ is one-one. If $f(x_2)<0$ then by continuity, there will be some $x_3$ between $x_1$ and $x_2$ such that $f(x_3)=0$. But that is again impossible due to $f(0)$ is already $0$ and $f$ is one-one. So $f(x_2)$ must be positive. This is true for all $x_2>0$.

If we don't assume $f$ is continuous then we can find counterexamples. For example $f(x)=x$ for $x$ rational, $f(x)=-x$ for $x$ irrational. This is symmetric around the origin, positive for some positive $x$ but not positive for all positive $x$.

Now let's consider the problem regarding the derivative. The derivative being positive for some $x>0$ does not imply it is positive for all $x$. We can flatten out the curve so that the derivative becomes $0$. For example $f(x)=x+\sin(x)$.

It is not true. Take, for instance$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\ne\pm1\\-x&\text{ if }x=1\text{ or }x=-1.\end{cases}\end{array}$$It is an odd function, and therefore it is symmetric about the origin. But $f(2)>0$ and $f(1)<0$.