Diagonalizing $xyz$ The quadratic form  $g(x,y) = xy$    can be diagonalized
by the change of variables  $x = (u + v)$   and  $y = (u - v)$ .
However, it seems unlikely that the cubic form  $f(x,y,z) = xyz$,
can be diagonalized by a linear change of variables.
Is there a short computational or theoretical proof of this?
Thanks.

If I understand correctly your question, you are asking if it is possible to write $xyz = \ell_1^3 + \ell_2^3 + \ell_3^3$ for some linear forms $\ell_1=\ell_1(x,y,z)$, $\ell_2,\ell_3$.
We can prove that $xyz \neq \ell_1^3 + \ell_2^3 + \ell_3^3$ in a few different ways. Here is a short theoretical proof that uses a little bit of projective geometry.
Lemma: If $xyz = \ell_1^3 + \ell_2^3 + \ell_3^3$, then $\{\ell_1,\ell_2,\ell_3\}$ are linearly independent.
Proof: The second derivatives of $xyz$ include $x$, $y$, and $z$. If $\ell_1,\ell_2,\ell_3$ span a space of dimension less than $3$, then they depend on only $2$ (or $1$) variables, and the second derivatives of each $\ell_i^3$ still only depend on $2$ (or $1$) variables, so they span less than $3$ dimensions. $\square$
Claim: $xyz \neq \ell_1^3 + \ell_2^3 + \ell_3^3$.
Proof: For convenience say $u=\ell_1$, $v=\ell_2$, $w=\ell_3$. The projective curve defined by $xyz=0$ is singular, in fact reducible. However the curve defined by $u^3+v^3+w^3=0$ is nonsingular. There is no linear change of coordinates that can carry a singular curve to a nonsingular one. $\square$
However, if you are not interested in projective geometry, it is still possible to deal with this via other approaches.
There is some literature on this subject under the name "Waring rank", Waring decompositions, symmetric tensors, and symmetric tensor rank. Some general introductions include Landsberg, Tensors: Geometry and Applications or Carlini, Grieve, Oeding, Four Lectures on Secant Varieties. That book and paper have basic explanations and references to further reading. I hope it helps.
