In the Munkres analysis on manifolds, after proving the implicit function theorem and saying that the choice of the last coordinates is given only for convenience, the following example:
suppose $A$ open in $R^5$ and $f:A \rightarrow R^2$ is a function of class $C^r$. Suppose one wishes to "solve" the equation $f(x,y,z,u,v)=0$ for the two unknowns $y$ and $u$ in terms of the other three. In this case, the implict function theorem tell us that if $a$ is a point of $A$ such that $f(a)=0$ and
$det\dfrac{\partial f}{\partial(y,u)}(a) \ne 0$,
then one can solve for $y$ and $u$ locally near that point, say $y= \phi (x,z,v)$ and $y= \psi (x,z,v)$. Furthermore, the derivatives of $\phi$ and $\psi$ satisfy the formula
$\dfrac{\partial (\phi,\psi)}{\partial(x,z,v)} = - \left[ \dfrac{\partial f}{\partial(y,u)} \right] ^{-1}.\dfrac{\partial f}{\partial(x,z,v)}$.
I'm not understanding how to get into this final formula, I've already tried to call $H (x,z,v) = (\phi,\psi)$, but it does not work. I can not get into this equation. Thanks for any tips.
Try to substitute $\phi$ and $\psi$ in $f$ and differentiate the result. You will get $$ 0 = \frac{\partial f(x,\phi(x,z,v),z,v, \psi(x,z,v))}{\partial (x,z,v)} = \frac{\partial f}{\partial (y,u)} \cdot \frac{\partial (\phi, \psi)}{\partial (x,z,v)} + \frac{\partial f}{\partial (x,z,v)}.$$ Then you can multiply this by $\left[ \dfrac{\partial f}{\partial(y,u)} \right] ^{-1}$ and get what you want.