Prove that there exist $a \in [-1,1]$, such that $f'''(a)=3f(1)-3f(-1)-6f'(0)$

I would use Taylor's theorem with Lagrange's form of remainder.

There exists $\xi_1\in$ a closed interval containing $a$ and $x$ such that:

$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2}+\frac{f'''(\xi_1)(x-a)^3}{6}$$

Now there exists an $a_1\in [0,1]$ such that $$f(1)=f(0)+f'(0)(1-0)+\frac{f''(0)(1-0)^2}{2}+\frac{f'''(a_1)(1-0)^3}{6} \tag{1}$$ Again there exists $a_2\in [-1,0]$ such that $$f(-1)=f(0)+f'(0)(-1-0)+\frac{f''(0)(-1-0)^2}{2}+\frac{f'''(a_2)(-1-0)^3}{6} \tag{2}$$

Subtract $(1)$ and $(2)$, we have $$3f(1)-3f(-1)-6f'(0)=\frac{f'''(a_1)+f'''(a_2)}{2}$$

We can prove that there exists an $a\in [a_1,a_2]$ such that $$f'''(a)=\frac{f'''(a_1)+f'''(a_2)}{2} \tag{3}$$

Rewriting we have a $a\in [a_1,a_2]$ such that $$f'''(a) = 3f(1)-3f(-1)-6f'(0)$$

(Use Darboux's property to prove $(3)$. As usual the hardest part is left to the reader! On a second thought, I thought I should write a hint. Assume $f'''(a_1)\le f'''(a_2)$ and observe $f'''(a_1)\le \frac{f'''(a_1)+f'''(a_2)}{2} \le {f'''}(a_2)$ and switch $a_1$ and $a_2$ to complete the proof)