Kernel of a mapping from a polynomial vector space to $\mathbb{R}^3$

It is almost correct, but the conclusion should be that $\ker f=\langle x^3-3x^2+2x\rangle$.

$\ker f=\operatorname{span}\{x(x-1)(x-2)\}$ has dimension $1$. Thus, by the Rank-nullity theorem, the image of $f$ has dimension $3$. So you are correct (other than losing an $x$ in $x^3-3x^2+2\color{red}x$...)