We need to check on which interval each function is increasing.

A function is increasing where its derivative is non-negative (and typically positive except possibly at isolated points).

1. Option A

Function: $f(x)=x^3-3x^2+3x+3$

Differentiate: $f'(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2$

Since $3(x-1)^2\ge 0\quad \text{for all }x\in(-\infty,\infty),$ $f(x)$ is increasing on all real numbers.

So A is correctly matched.

2. Option B

Function: $f(x)=2x^3-3x^2-12x+6$

Differentiate: $f'(x)=6x^2-6x-12=6(x^2-x-2)=6(x-2)(x+1)$

For the interval $[2,\infty)$:

• if $x=2$, then $f'(2)=0$
• if $x>2$, then both $(x-2)>0$ and $(x+1)>0$, so $f'(x)>0$

Hence $f(x)$ is increasing on $[2,\infty)$.

So B is correctly matched.

3. Option C

Function: $f(x)=3x^2-2x+1$

Differentiate: $f'(x)=6x-2$

For increasing, we need: $6x-2\ge 0 \implies x\ge \frac13$

So the function is increasing on $\left[\frac13,\infty\right)$

But the given interval is $\left(-\infty,\frac13\right]$ which is actually the interval where the function is decreasing.

So C is incorrectly matched.

4. Option D

Function: $f(x)=x^3+6x^2+6$

Differentiate: $f'(x)=3x^2+12x=3x(x+4)$

Now check the interval $(-\infty,-4)$:

• for $x<-4$, we have $x<0$ and $x+4<0$
• so $x(x+4)>0$
• hence $f'(x)>0$

Therefore the function is increasing on $(-\infty,-4)$.

So D is correctly matched.

Final conclusion

The incorrectly matched pair is: $\boxed{\text{C}}$

This matches the stored correct answer.