A global version of the implicit function theorem

[wrobel]

I do not know if this construction is contained in textbooks but it is surely contained in the collective unconscious :)

Let $D\subset\mathbb{R}^s$ be an open set such that for some fixed point $x_0\in D$ and for any $x\in D$ the interval $[x_0,x]$ belongs to $D.$

Consider a mapping $f\in C^2(D\times\mathbb{R}^m,\mathbb{R}^n),\quad f=f(x,y),\quad n\le m$ such that for some $y_0\in\mathbb{R}^m$ one has
$$f(x_0,y_0)=0;$$ and for any $(x,y)\in D\times \mathbb{R}^m$
it follows that $$\mathrm{rang}\,\frac{\partial f}{\partial y}=n.$$

Introduce a matrix
$$A(x,y)=-\Big(\frac{\partial f}{\partial y}\Big)^T\Big(\frac{\partial f}{\partial y}\Big(\frac{\partial f}{\partial y}\Big)^T\Big)^{-1}\frac{\partial f}{\partial x}.$$

Theorem. Assume that
$$\sup\{\|A(x,y)\|:\quad (x,y)\in D\times\mathbb{R}^m\}<c_1\|y\|+c_2.$$
Then the equation
$$f(x,y)=0$$ has a solution $y=y(x)\in C^1(D,\mathbb{R}^m)$ such that
$y(x_0)=y_0.$

Indeed, let $u(t,x)$ be a solution to the following Cauchy problem
$$u_t=A((1-t)x_0+tx,u)\cdot (x-x_0),\quad u\mid_{t=0}=y_0.\qquad(1)$$
It is easy to see that $u$ is defined for $t\in[0,1]$ and $y(x)=u(1,x).$

Hint: $u_t$ from equation (1) solves the equation
$$\frac{d}{dt}f((1-t)x_0+tx,u)=0.$$