For the function $h(x)=\frac{6x}{(x+1)(x-2)}$, solve the following inequality.
$$h(x)<0$$
Select the correct choice below and fill in the answer box within your choice.

Step 1 ：The inequality is $h(x)<0$, where $h(x)=\frac{6x}{(x+1)(x-2)}$.

Step 2 ：To solve this inequality, we need to find the values of $x$ for which $h(x)<0$.

Step 3 ：First, we find the values of $x$ for which $h(x)=0$. This happens when the numerator of $h(x)$ is zero, i.e., when $x=0$.

Step 4 ：Next, we find the values of $x$ for which $h(x)$ is undefined. This happens when the denominator of $h(x)$ is zero, i.e., when $x=-1$ or $x=2$.

Step 5 ：Now, we consider the intervals $(-\mathrm{\infty },-1)$, $(-1,0)$, $(0,2)$, and $(2,\mathrm{\infty })$.

Step 6 ：In each interval, we choose a test point and substitute it into $h(x)$ to determine whether $h(x)<0$ in that interval.

Step 7 ：For $(-\mathrm{\infty },-1)$, we choose $x=-2$. Substituting $x=-2$ into $h(x)$, we get $h(-2)=\frac{6(-2)}{((-2)+1)((-2)-2)}=\frac{-12}{(-1)(-4)}=-3<0$. So, $h(x)<0$ in $(-\mathrm{\infty },-1)$.

Step 8 ：For $(-1,0)$, we choose $x=-0.5$. Substituting $x=-0.5$ into $h(x)$, we get $h(-0.5)=\frac{6(-0.5)}{((-0.5)+1)((-0.5)-2)}=\frac{-3}{0.5(-2.5)}=1.2>0$. So, $h(x)<0$ does not hold in $(-1,0)$.

Step 9 ：For $(0,2)$, we choose $x=1$. Substituting $x=1$ into $h(x)$, we get $h(1)=\frac{6(1)}{(1+1)(1-2)}=\frac{6}{-1}=-6<0$. So, $h(x)<0$ in $(0,2)$.

Step 10 ：For $(2,\mathrm{\infty })$, we choose $x=3$. Substituting $x=3$ into $h(x)$, we get $h(3)=\frac{6(3)}{(3+1)(3-2)}=\frac{18}{2}=9>0$. So, $h(x)<0$ does not hold in $(2,\mathrm{\infty })$.

Step 11 ：Final Answer: The solution to the inequality $h(x)<0$ is $\boxed{{\displaystyle (-\mathrm{\infty },-1)\cup (0,2)}}$.