A nilpotent endomorphism

Linearity. Each coordinate of $f(x,y,z)$ is a linear form in $(x,y,z)$ (namely $y$, $z$, $0$), so $f$ is linear.

Powers. $f^2(x,y,z)=f(y,z,0)=(z,0,0)$ and $f^3(x,y,z)=f(z,0,0)=(0,0,0)$. So $f^3=0$ while $f^2\neq0$: $f$ is nilpotent of index exactly $3$.

Kernel. $f(x,y,z)=0\iff y=z=0$, so $\ker f=\operatorname{span}\big((1,0,0)\big)$, of dimension $1$.

Image. $\operatorname{Im} f=\{(y,z,0)\}=\operatorname{span}\big((1,0,0),(0,1,0)\big)$, of dimension $2$.

Check. $\dim\ker f+\dim\operatorname{Im} f=1+2=3$.

Conclusion. $f$ is nilpotent of index $3$, and $\ker f=\operatorname{span}(e_1)\subsetneq\operatorname{Im} f=\operatorname{span}(e_1,e_2)$. In particular $\ker f\cap\operatorname{Im} f\neq\{0\}$.