Let $h(x)=f(x)-3g(x).$ Then $h$ is continuous on $[-1,2]$ and twice differentiable on $(-1,2)$.

We are given: $f(-1)=3,\, f(0)=6,\, f(2)=0$ $g(-1)=0,\, g(0)=1,\, g(2)=-1$ So, $h(-1)=f(-1)-3g(-1)=3-0=3,$ $h(0)=f(0)-3g(0)=6-3=3,$ $h(2)=f(2)-3g(2)=0-3(-1)=3.$ Hence $h(-1)=h(0)=h(2)=3.$

Also, in each of the intervals $(-1,0)$ and $(0,2)$, we are told that $(f-3g)''=h''$ never vanishes. That means $h''(x)\neq 0$ throughout each of these two intervals.

1. Existence of a zero of $h'$ in $(-1,0)$

Since $h(-1)=h(0),$ by Rolle's theorem there exists at least one $c_1\in(-1,0)$ such that $h'(c_1)=0.$ That is, $f'(c_1)-3g'(c_1)=0.$

So there is at least one solution in $(-1,0)$.

2. Existence of a zero of $h'$ in $(0,2)$

Similarly, since $h(0)=h(2),$ by Rolle's theorem there exists at least one $c_2\in(0,2)$ such that $h'(c_2)=0.$

So there is at least one solution in $(0,2)$.

3. Uniqueness in each interval using $h''\neq 0$

Since $h''(x)\neq 0$ on $(-1,0)$ and $h''$ is continuous there, $h''$ cannot change sign on $(-1,0)$. Thus on $(-1,0)$ either $h''(x)>0 \quad \text{for all }x\in(-1,0),$ or $h''(x)<0 \quad \text{for all }x\in(-1,0).$ Therefore $h'$ is strictly monotonic on $(-1,0)$. A strictly monotonic function can have at most one zero.

But from Rolle's theorem, it has at least one zero. Hence it has exactly one zero in $(-1,0)$.

The same argument applies on $(0,2)$:

• $h''(x)\neq 0$ and continuous on $(0,2)$,
• so $h''$ has constant sign there,
• hence $h'$ is strictly monotonic on $(0,2)$,
• therefore $h'$ can have at most one zero there,
• and by Rolle's theorem it has at least one zero there.

So $h'$ has exactly one zero in $(0,2)$.

Thus, $f'(x)-3g'(x)=0$ has exactly one solution in $(-1,0)$ and exactly one solution in $(0,2)$.

4. Checking the options

Option A

Claims exactly three solutions in $(-1,0)\cup(0,2)$.

But we found exactly one in each interval, so total exactly two solutions. Hence A is false.

Option B

Claims exactly one solution in $(-1,0)$. This is true. Hence B is true.

Option C

Claims exactly one solution in $(0,2)$. This is true. Hence C is true.

Option D

Claims exactly two solutions in $(-1,0)$ and exactly two in $(0,2)$. This contradicts strict monotonicity of $h'$ on each interval. Hence D is false.

Final Answer

The correct statements are: $\boxed{B,\ C}$