How to check the independence of functions of some variables? Given some functions $f(x,y,z),\  g(x,y,z),\, $ and $h(x,y,z)$, can I check their independence by the following criterion? 

$f,g$ and $h$ are independent if their gradients (which now becomes a vector with the bases $\hat{x},\hat{y},\hat{z}$) are linearly independent. 

So, now the problem becomes that of linear algebra.
Is the above line of action correct? If the above statement is correct, please also refer me to a proof.
 A: As I said in the comments, if $\nabla f, \nabla g, \nabla h$ are linearly independent, then so are $f, g, h$. In particular, if we have scalars, $a, b, c$ such that
$$af + bg + ch = 0$$
as functions, then taking $\nabla$ of both sides yields
$$a \nabla f + b \nabla g + c \nabla h = \nabla 0 = 0.$$
If $\nabla f, \nabla g, \nabla h$ are linearly independent, then $a = b = c = 0$, and hence $f, g, h$ are linearly independent.
The question is, does the converse hold? If $f, g, h$ are linearly independent, is the same true for $\nabla f, \nabla g, \nabla h$? Unfortunately, the converse doesn't hold in general. I've adapted the corresponding counterexample that Peano provided for Wronskians. Let:
\begin{align*}
f(x, y, z) &= x^2y^2z^2 \\
g(x, y, z) &= x^2y^2z|z| \\
h(x, y, z) &= x^2yz|yz|.
\end{align*}
Then
\begin{align*}
\nabla f(x, y, z) &= (2xy^2z^2, 2x^2yz^2, 2x^2y^2z) \\
\nabla g(x, y, z) &= (2xy^2z|z|, 2x^2yz|z|, 2x^2y^2|z|) \\
\nabla h(x, y, z) &= (2xyz|yz|, 2x^2z|yz|, 2x^2y|yz|).
\end{align*}
Putting these as columns in a matrix,
$$A = \begin{bmatrix} 2xy^2z^2 & 2xy^2z|z| & 2xyz|yz| \\ 2x^2yz^2 & 2x^2yz|z| & 2x^2z|yz| \\ 2x^2y^2z & 2x^2y^2|z| & 2x^2y|yz| \end{bmatrix}.$$
Using the multilinearity of the determinant (as a function of column vectors), we can pull some factors out of the determinant:
$$\det A = (2xyz) \cdot (2xy|z|) \cdot (2x|yz|) \cdot \det \begin{bmatrix} yz & yz & yz \\ xz & xz & xz \\ xy & xy & xy \end{bmatrix} = 0.$$
However, I claim that $f, g, h$ are linearly independent. Suppose that $a, b, c$ exist such that
$$af + bg + ch = 0$$
which is to say, for all $x, y, z \in \Bbb{R}$,
$$ax^2 y^2 z^2 + b x^2 y^2 z |z| + c x^2 yz |yz| = 0.$$
For example, if $(x, y, z) = (1, 1, 1)$,
$$a + b + c = 0.$$
Another example: if $(x, y, z) = (1, 1, -1)$, then
$$a - b - c = 0.$$
Similarly, if $(x, y, z) = (1, -1, -1)$, then
$$a - b + c = 0.$$
The only possible values for $a, b, c$ is $a = b = c = 0$, hence $f, g, h$ are linearly independent, completing the counterexample.
