In Advanced Calculus by Widder I have come across a problem on page 30 regarding Cramer's rule and solving for a partial derivative.
I understand the following:
\begin{align} F\left(u,v,g\left(u,v,x\right)\right)&=0,\tag{1}\\ G\left(u,v,h\left(u,v,y\right)\right)&=0,\tag{2} \end{align} so in differentiating $\displaystyle\frac{\partial u}{\partial y}$ we arive at \begin{align} \begin{cases} \displaystyle\frac{\partial F}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial g}\left[\frac{\partial g}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial y}+\require{cancel}\xcancel{\frac{\partial g}{\partial x}\frac{\partial x}{\partial y}}\right]=0\\ \displaystyle\frac{\partial G}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial G}{\partial h}\left[\frac{\partial h}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial h}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial h}{\partial y}\require{cancel}\xcancel{\frac{\partial y}{\partial y}}\right]=0 \end{cases}\tag{3}\\ \implies\begin{cases} \displaystyle\frac{\partial F}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial g}\left[\frac{\partial g}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial y}\right]=0,\\ \displaystyle\frac{\partial G}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial G}{\partial h}\left[\frac{\partial h}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial h}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial h}{\partial y}\right]=0. \end{cases}\tag{4} \end{align} This makes total sense as it's really just an extension of the chain rule. But now, how do I solve for $\displaystyle \frac{\partial u}{\partial y}$ using $\left(4\right)$ above? In other words, what will the coefficients be in the determinants given that I now have a chain rule? Why will they be the partial derivatives of $F$ and $G$ as he has written on the next page to give us \begin{align} \begin{vmatrix} \displaystyle\frac{\partial F}{\partial u}+\frac{\partial F}{\partial g}\frac{\partial g}{\partial u} & \displaystyle\frac{\partial F}{\partial v}+\frac{\partial F}{\partial g}\frac{\partial g}{\partial v}\\ \displaystyle\frac{\partial G}{\partial u}+\frac{\partial G}{\partial h}\frac{\partial h}{\partial u} & \displaystyle\frac{\partial G}{\partial v}+\frac{\partial G}{\partial h}\frac{\partial h}{\partial v} \end{vmatrix},\tag{5} \end{align} which like the Wronskian, $\neq 0$. Is it always just the coefficients out front of any partial derivative or term with respect to $y$?