f(x) = 0 has min two roots in (0, 2)

$\Rightarrow f^{\mathrm{\prime }}\left(0\right)=0$ has min 1 root in (0, 2)

$\therefore f\left(2\right)=f\left(3\right)=-2$

$\therefore f^{\mathrm{\prime }}\left(x\right)=0 \text{has min}1 \text{root in }(2,3)$

$\therefore f^{\mathrm{\prime }}\left(x\right)=0 \text{has}min2\text{root in }(0,3)$

$h^{\mathrm{\prime }}\left(x\right)={\left(f^{\mathrm{\prime }2}\right(x)\cdot f(x\left)\right)}^{\mathrm{\prime }}=2f^{\mathrm{\prime }}\left(x\right)\cdot f^{\prime \prime }\left(x\right)\cdot f\left(x\right)+f^3\left(x\right)$

$=0 \text{has min }3 \text{roots in }(0,3)$