For the function $h(x)=\frac{6 x}{(x+1)(x-2)}$, 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 \(h(x)=\frac{6 x}{(x+1)(x-2)}\).

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 \((-\infty,-1)\), \((-1,0)\), \((0,2)\), and \((2,\infty)\).

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 \((-\infty,-1)\), 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 \((-\infty,-1)\).

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 \((2,\infty)\), 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 \((2,\infty)\).

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