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Let $U_1\subset \mathbb{R}^k$ and $U_2\subset \mathbb{R}^m$ be open subsets and $$F:U_1\times U_2\mathbb{R}^m\\ (x,y)\mapsto (F_1(x,y), \ldots , F_m(x,y))$$ be a differentiable function. Let $g:U_1\rightarrow U_2$ be a differentiable function such that $F(x,g(x))=0$ for all $x\in U_1$.

We suppose that the $(m\times m)$-Matrix $D_yF(x_0, y_0)$ in a point $(x_0, g(x_0)):=(x_0,y_0)$ with $x_0\in U_1$ is invertible.

Show that $$Dg(x_0)=-(D_yF(x_0,y_0))^{-1}(D_xF(x_0,y_0))$$

Does this mean that we have to give the proof of Implicit function theorem?

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No, you don't have to prove the implicit function theorem.

You only need to calculate

$$D_xF(x,g(x))$$

and then solve it for $D_xg(x)$.

To do so, just use the $\color{blue}{\text{chain rule}}$ for total derivatives:

$$D_xF(x,g(x)) = \left. D_xF(x,y)\right|_{y=g(x)} + \color{blue}{\left. D_yF(x,y)\right|_{y=g(x)}D_xg(x)} = 0_{m\times k}$$

Note that $\left.D_yF(x,y)\right|_{y=g(x)}$ is invertible at $(x_0,y_0) = (x_0, g(x_0))$. So, you can solve above equation for $D_xg(x_0)$ and get the required expression.

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