Let \(k,n\geq1\) and \(r\geq1\), let \(A\subseteq\RR^{k+n}\) be open, and let \(f\colon A\to\RR^n\) be of class \(C^r\). Write \(f(\vector x,\vector y)\) with \(\vector x\in\RR^k\) and \(\vector y\in\RR^n\), so that \[ Df=\begin{bmatrix}\dfrac{\partial f}{\partial\vector x} & \dfrac{\partial f}{\partial\vector y}\end{bmatrix} \] splits into an \(n\times k\) block and an \(n\times n\) block. Suppose \((\vector a,\vector b)\in A\) satisfies \(f(\vector a,\vector b)=0\) and \[ \det\frac{\partial f}{\partial\vector y}(\vector a,\vector b)\neq0. \] Then there are a neighbourhood \(B\subseteq\RR^k\) of \(\vector a\) and a unique continuous \(g\colon B\to\RR^n\) with \(g(\vector a)=\vector b\) and \[ f(\vector x,g(\vector x))=0\quad\text{for all }\vector x\in B, \] and this \(g\) is of class \(C^r\).
Differentiating \(f(\vector x,g(\vector x))=0\) gives \[ \frac{\partial f}{\partial\vector x}+\frac{\partial f}{\partial\vector y}\,Dg=0, \qquad\text{so}\qquad Dg(\vector x)=-\Bigl[\frac{\partial f}{\partial\vector y}(\vector x,g(\vector x))\Bigr]^{-1}\frac{\partial f}{\partial\vector x}(\vector x,g(\vector x)). \] The proof applies the inverse function theorem to \(F(\vector x,\vector y)\coloneqq(\vector x,f(\vector x,\vector y))\), whose derivative at a point is invertible if and only if \(\partial f/\partial\vector y\) is invertible there.
See [@Mun91, Section 9, Theorem 9.2]; the formula for \(Dg\) is [@Mun91, Theorem 9.1].