# Is this the proof of Implicit function theorem?

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?

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.