Suppose
\[f(x)=x+y+z\]
, \[g(x)=x^2 +y+z\]
, \[h(x)=x^3 +y+z\]
Then
\[\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=1\]
, \[\frac{\partial f}{\partial x}=2x, \frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=1\]
, \[\frac{\partial f}{\partial x}=3x^2 , \frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=1\]
The matrix of partial derivatives is
\[ \left| \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} & \frac{\partial h}{\partial x} \\ \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y} & \frac{\partial h}{\partial y} \\ \frac{\partial f}{\partial z} & \frac{\partial g}{\partial z} & \frac{\partial h}{\partial z} \end{array} \right| = \left| \begin{array}{ccc} 1 & 2x & 3x^2 \\ 1 & 1 & 1 \\ 1 & 1 & \1 \end{array} \right| =0\]
since the second and third rows are the same. This means we can dind a function
\[F\]
satisfying \[F(f,g,h)=0\]
.We can treat the problem as a transformation from the
\[(x,y,z)\]
space to the \[(f,g,h)\]
plane. The transformation is degenerate.