# multivariable implicit function theorem

I'm trying to prove that if $f(x_0)=0$ and $\det \begin{bmatrix} \frac{\partial f_1(x_0)}{\partial x_1} & \frac{\partial f_1 (x_0)}{\partial x_3} \\ \frac{\partial f_2 (x_0)}{\partial x_1} & \frac{\partial f_2(x_0)}{\partial x_3} \end{bmatrix} \neq 0$, where $f\in C^1(\Omega,\mathbb{R}^2), \Omega$ open subset of $\mathbb{R^2}$, then exists $\phi:[x_{02}-\varepsilon,x_{02}+\varepsilon]\to [x_{01}-\delta,x_{01}+\delta]\times [x_{03}-h,x_{03}+h]$ such that $$\phi_1'(y)=-\dfrac{\det\begin{bmatrix} \frac{\partial f_1(\phi_1(y),y,\phi_2(y))}{\partial x_2} & \frac{\partial f_1 (\phi_1(y),y,\phi_2(y)}{\partial x_3} \\ \frac{\partial f_2 (\phi_1(y),y,\phi_2(y))}{\partial x_2} & \frac{\partial f_2(\phi_1(y),y,\phi_2(y))}{\partial x_3} \end{bmatrix} }{\det\begin{bmatrix} \frac{\partial f_1(\phi_1(y),y,\phi_2(y))}{\partial x_1} & \frac{\partial f_1 (\phi_1(y),y,\phi_2(y))}{\partial x_3} \\ \frac{\partial f_2 (\phi_1(y),y,\phi_2(y))}{\partial x_1} & \frac{\partial f_2(\phi_1(y),y,\phi_2(y))}{\partial x_3} \end{bmatrix} }$$ and $$\phi_2'(y)=-\dfrac{\det\begin{bmatrix} \frac{\partial f_1(\phi_1(y),y,\phi_2(y))}{\partial x_1} & \frac{\partial f_1 (\phi_1(y),y,\phi_2(y)}{\partial x_2} \\ \frac{\partial f_2 (\phi_1(y),y,\phi_2(y))}{\partial x_2} & \frac{\partial f_2(\phi_1(y),y,\phi_2(y))}{\partial x_2} \end{bmatrix} }{\det\begin{bmatrix} \frac{\partial f_1(\phi_1(y),y,\phi_2(y))}{\partial x_1} & \frac{\partial f_1 (\phi_1(y),y,\phi_2(y))}{\partial x_3} \\ \frac{\partial f_2 (\phi_1(y),y,\phi_2(y))}{\partial x_1} & \frac{\partial f_2(\phi_1(y),y,\phi_2(y))}{\partial x_3} \end{bmatrix} }.$$ To me it seems just a generalization of the classic Implicit Function theorem with $f:\Omega\subset\mathbb{R}^2\to\mathbb{R}$ but I don't know how to got started with a proof of this result so any help would be appreciated.

Thanks.

• You can find proofs of this theorem in just about any analysis textbook. It is a generalization of what you call the classic implicit function theorem, but that one has at least one easier proof that does not work in higher dimensions. – Lukas Geyer Jun 21 '18 at 0:37