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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##
the following non degenerateness condition holds $$\mathrm{rang}\,\frac{\partial f}{\partial y}=n.$$
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
$$\frac{d}{dt}f((1-t)x_0+tx,u)=0.$$
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##
the following non degenerateness condition holds $$\mathrm{rang}\,\frac{\partial f}{\partial y}=n.$$
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
$$\frac{d}{dt}f((1-t)x_0+tx,u)=0.$$
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