Let $U \subseteq \mathbb{R}^2$ be an open neighborhood of $(0,0) \in \mathbb{R}^2$. Let $F: U \to \mathbb{R}$ be a three times continuous differentiable function with $D_2F(0,0) \neq 0$. It exists a function $g:(- \varepsilon, \varepsilon) \to \mathbb{R}$ for a $\varepsilon > 0$ with $g(0) = 0$ and $F(x,g(x)) = 0$ for all $x \in (- \varepsilon, \varepsilon)$.
Now I have to show that $g$ is three times continuous differentiable in a neigborhood of $0 \in \mathbb{R}$ and I have to find formulas for $g', g''$ and $g'''$ that depend only on $g$ and the derivatives of $F$.
The last part should be easy, since we can write $g'(x)=- \frac{D_1F(x,g(x))}{D_2F(x,g(x))}$. $g''$ and $g'''$ can be calculated by differentiation of the previous derivative.
So my question is how to prove that $g$ is three times continuous differentiable in a neighborhood of $0 \in \mathbb{R}?$ I would appreciate some hints/ help.
Differentiability of $g$ in a neighbourhood of $0$ is one of the implications of the implicit function theorem.
So you can differentiate $F(x,g(x)) = $ in a whole neighbourhood of $x=0$:
$$F_1(x,g(x)) =- F_2(x,g(x)) g^\prime(x)$$ By assumption and continuity of $D_2F$ and $g(x)$, $D_2F(x,g(x)) \neq 0$ in a neighbourhood of $x=0$, so this can be solved for $g^\prime(x)$ in a neighbourhood of $0$. If you do that you will arrive at an equation $g^\prime(x) = $ something of class $C^1$ (it's not difficult, you already did it, but I want to leave something to do for you...).
So $g$ is $C^2 $ in a neighbourhood of $0$. Differentiate once more to conclude it's even $C^3$.