Answer
$\text{No}$
Work Step by Step
We know that a function is odd if and only if $f(-x)=-f(x)$ and a function is one one if $x_{1}= x_{2}$ then $f(x_{1})=f(x_{2})$.
Let, $f(x)=x^{3}-x$ then clearly $f(-x)= -f(x)$. Hence $f(x)$ is odd function.
Let $x_{1}= -1$ & $x_{2}=1$ then
$f(x_{1})= -1+1= 0$
$f(x_{2})= 1-1=0$
Here, we have $f(x_{1})= f(x_{2})$ but $x_{1}\ne x_{2}$
Hence $f$ is not one one.
Therefore not all odd functions are one-to-one.
