How to use the second derivative test to find all relative extrema of f (...

The first derivative is ${\displaystyle f\prime \left(x\right)=6x-3x^2=3x(2-x)}$, which has roots at ${\displaystyle x=0}$ and ${\displaystyle x=2}$. These are the turning point as well as the potential sites of local extrema.
Since the second derivative is ${\displaystyle f\prime \prime \left(x\right)=6-6x}$, we get ${\displaystyle f\prime \prime \left(0\right)=6>0}$ and ${\displaystyle f\prime \prime \left(2\right)=-6<0}$. The fact that ${\displaystyle f\prime \prime \left(0\right)>0}$ (and the fact that ${\displaystyle f\prime \prime }$ is continuous) implies that the graph of ${\displaystyle f}$ is concave up near ${\displaystyle x=0}$, making, by the Second Derivative Test, ${\displaystyle x=0}$ the location of a local minimum.
The fact that ${\displaystyle f\prime \prime \left(2\right)<0}$ (and the fact that ${\displaystyle f\prime \prime }$ is continuous) implies that the graph of ${\displaystyle f}$ is concave down near ${\displaystyle x=2}$, making, by the Second Derivative Test, ${\displaystyle x=2}$ the location of a local maximum.
The local minimum value (output) is ${\displaystyle f\left(0\right)=5}$ and the local maximum value (output) is ${\displaystyle f\left(2\right)=5+12-8=9}$.