Prove $f$ is derivable and find $f'(0)$ I'm stuck solving the following problem:

Problem: Prove there exists an unique function $f : \mathbb{R} \to \mathbb{R}$ such that:
$$2(f(x))^3 - 3(f(x))^2 + 6f(x) = x \quad \forall x \in \mathbb{R}$$
Also prove $f$ is derivable in $\mathbb{R}$ and calculate $f'(0)$.

I have been able to prove the existence and uniqueness (see below) but I don't know how to prove $f$ is derivable or calculate $f'(0)$. 
How can I prove it?

Proof of existence and uniqueness:
Let $a \in \mathbb{R}$. The equation 
$$2x^3 - 3x^2 + 6x = a \qquad (\ast)$$
has at least one solution since $p(x) = 2x^3 - 3x^2 + 6x$ is an odd-degree polynomial. 
Suppose there were two different solutions $x_1, x_2$ for $(\ast)$. Then, by Rolle's theorem there exists $c \in (x_1, x_2)$ such that:
$$p'(c) = 6x^2 - 6x + 6 = 0$$
But $\Delta_{p'} = 6^2 - 4 \cdot 6^2 < 0$, so $p'(c) = 0$ has no real solutions, which is a contradiction. Hence, there exists an unique solution for $(\ast)$.
Finally, we can define $f(a) = x$, where $x$ is the solution to $(\ast)$.
 A: Set $p(f)=2f^3-3f^2+6f$ since $p'(f)>0$ then by Inverse function theorem $p$ is invertible and $p^{-1}$ is differentiable and we have:
$$f(x)=p^{-1} (x)$$
For calculate $f'$, we derive from both sides of $2(f(x))^3 - 3(f(x))^2 + 6f(x) = x$ then we achieve this equation:
$$f'(x)=\frac{1}{6(f(x)^2 - f(x)+1)}$$ 
If  $z$ is real root of $p(f)=0$ then $f'(0)=\frac{1}{6(z^2-z+1)}$ and because $z=0$ then $f'(0)=\frac{1}{6}$.  

Generalization:
  If $g(f(x))$ and $h(x)$ are real functions such that $g'(f(x))>0$ and $ h'(x)>0 \quad \forall x \in \mathbb R$ now if we have $g(f(x))=h(x)$ then $f(x)$ is differentiable well-defined function.

A: Uniqueness comes from, for any $x$,
$$2f^3-3f^2+6f=2g^3-3g^2+6g,$$or
$$(f-g)\left(2(f^2+fg+g^2)-3(f+g)+6\right)=0.$$
Solving $2f^2+(2g-3)f+2g^2-3g+6$ for $f$, the discriminant is
$$(2g-3)^2-4(2g^2-3g+6)=-4g^2-15,$$
so that there is no other real solution than $f=g$.

For $x=0$, $2f^3-3f^2+6f=f(2f^2-3f+6)=0$ has the single real solution $f=0$.
Then $f'\left(6f^2-6f+6\right)=1$, and $f'(0)=\dfrac16.$
As $6f^2-6f+6$ has no real roots, the derivative is always defined.
