$\text{The function}f\left(x\right)=\frac{1}{12}(3x^4+4x^3-12x^2)\text{ decreases in}$

The correct option is:
$(-\infty ,-2)\cup (0,1)$

Step-by-step Explanation:

A function $f\left(x\right)$ is strictly decreasing on an interval where its first derivative is negative, i.e., $f^{\prime }\left(x\right)<0$.

First, let us write down the given function:
$f\left(x\right)=\frac{1}{12}(3x^4+4x^3-12x^2)$

Now, we find the first derivative of the function with respect to $x$:
$f^{\prime }\left(x\right)=\frac{1}{12}\frac{d}{dx}(3x^4+4x^3-12x^2)$
Using the power rule for differentiation:
$f^{\prime }\left(x\right)=\frac{1}{12}(12x^3+12x^2-24x)$
Simplifying the expression by dividing each term inside the parenthesis by 12:
$f^{\prime }\left(x\right)=x^3+x^2-2x$

To analyze the intervals of decrease, we factor the derivative:
$f^{\prime }\left(x\right)=x(x^2+x-2)$
Factoring the quadratic term $x^2+x-2$:
$x^2+x-2=(x+2)(x-1)$
Thus, the fully factored derivative is:
$f^{\prime }\left(x\right)=x(x+2)(x-1)$

Next, we determine the critical points where $f^{\prime }\left(x\right)=0$:
$x=-2,x=0,\text{and}x=1$

These critical points divide the real line into four intervals:
1. $(-\infty ,-2)$
2. $(-2,0)$
3. $(0,1)$
4. $(1,\infty )$

We test the sign of $f^{\prime }\left(x\right)$ in each of these intervals:

- For the interval $(-\infty ,-2)$, choose $x=-3$:
$f^{\prime }(-3)=(-3)(-3+2)(-3-1)=(-3)(-1)(-4)=-12<0$ (Decreasing)

- For the interval $(-2,0)$, choose $x=-1$:
$f^{\prime }(-1)=(-1)(-1+2)(-1-1)=(-1)\left(1\right)(-2)=2>0$ (Increasing)

- For the interval $(0,1)$, choose $x=0.5$:
$f^{\prime }\left(0.5\right)=\left(0.5\right)(0.5+2)(0.5-1)=\left(0.5\right)\left(2.5\right)(-0.5)=-0.625<0$ (Decreasing)

- For the interval $(1,\infty )$, choose $x=2$:
$f^{\prime }\left(2\right)=\left(2\right)(2+2)(2-1)=\left(2\right)\left(4\right)\left(1\right)=8>0$ (Increasing)

Comparing the signs, the derivative $f^{\prime }\left(x\right)$ is negative in the intervals:
$(-\infty ,-2)\cup (0,1)$

Therefore, the function decreases in:
$(-\infty ,-2)\cup (0,1)$