1. We need to determine where $f(x)=3\sin^4 x+10\sin^3 x+6\sin^2 x-3$ is increasing or decreasing on the given intervals.

2. Differentiate $f(x)$: $f'(x)=12\sin^3 x\cos x+30\sin^2 x\cos x+12\sin x\cos x$ Factor it: $f'(x)=6\sin x\cos x\left(2\sin^2 x+5\sin x+2\right)$ Now factor the quadratic: $2\sin^2 x+5\sin x+2=(2\sin x+1)(\sin x+2)$ So, $f'(x)=6\sin x\cos x(2\sin x+1)(\sin x+2)$

3. Now analyze the sign of each factor on the domain $x\in\left[-\frac\pi6,\frac\pi2\right].$

• On this interval, $\sin x\in\left[-\frac12,1\right]$ So, $\sin x+2>0$ throughout the interval.

• Also, on $\left(-\frac\pi6,\frac\pi2\right),$ we have $\cos x>0.$

Thus the sign of $f'(x)$ depends on $\sin x(2\sin x+1).$

4. Check interval $\left(-\frac\pi6,0\right)$:

Here, $\sin x\in\left(-\frac12,0\right).$ So,

• $\sin x<0$
• $2\sin x+1>0$
• $\cos x>0$
• $\sin x+2>0$

Hence, $f'(x)<0$ on $\left(-\frac\pi6,0\right)$. Therefore, $f$ is decreasing on $\left(-\frac\pi6,0\right).$ So option $D$ is correct, and option $C$ is false.

5. Check interval $\left(0,\frac\pi2\right)$:

Here, $\sin x>0,\quad \cos x>0,\quad 2\sin x+1>0,\quad \sin x+2>0$ Therefore, $f'(x)>0$ on $\left(0,\frac\pi2\right)$. So $f$ is increasing there. Hence option $B$ is false.

6. Check option $A$: increasing on the whole interval $\left(-\frac\pi6,\frac\pi2\right).$ This is false because:

• $f'(x)<0$ on $\left(-\frac\pi6,0\right)$
• $f'(x)>0$ on $\left(0,\frac\pi2\right)$ So it is not increasing on the entire interval.

7. Final conclusion: Only option $D$ is correct.