Exercise 9.19

Let

Then the matrix of ${\check{f}}^{\prime }(x,y,z,u)$ is

Note that $\check{f}(\check{0})=\check{0}$. The $x,y,u$ part of ${\check{f}}^{\prime }$ has determinant $8u-12\ne 0$ near $\check{0}$, so by the implicit function theorem, there is a solution of $\check{f}(x(z),y(z),z,u(z))=\check{0}$ near $\check{0}$.

Similarly, the determinant of the $x,z,u$ part of ${\check{f}}^{\prime }$ is equal to $21-14u\ne 0$ near $\check{0}$, so there is a solution of $\check{f}(x(y),y,z(y),u(y))=\check{0}$ near $\check{0}$. And the determinant of the $y,z,u$ part of ${\check{f}}^{\prime }$ is equal to $3-2u\ne 0$ near $\check{0}$, so there is a solution of $\check{f}(x,y(x),z(x),u(x))=\check{0}$ near $\check{0}$.

However, the determinant of the $x,y,z$ part of ${\check{f}}^{\prime }$ is equal to 0, so the implicit function theorem cannot be applied. If you try to solve for $x,y,z$ in terms of $u$, you just get an equation in $u$ which has no solution near $\check{0}$.