I begin my question with a multilinear question then I will consider two local smooth analogies:


Assume that $\alpha$ is a real valued symmetric $k$-tensor, that is a $k$-linear map $\alpha:\overbrace{\mathbb{R}^n\times\mathbb{R}^n\times\ldots\times \mathbb{R}^n}^{k\; \text{times}} \to \mathbb{R}$ which is unchanged under permutations.



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> Is there an (orthonormal) linear map $T$ on $\mathbb{R}^n$ such that $T^\*(\alpha)$ is in the diagonal form $T^\*(\alpha)=\sum\_{i=1}^n \lambda\_i \overbrace{dx\_i\otimes dx\_i\otimes \ldots \otimes dx\_i}^{k \;\text{times}}$
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We would like to consider two kind of smooth analogies of this multilinear question as follows:


Assume that $\omega$ is a smooth symmetric $k$ tensor on an open set of $\mathbb{R}^n$:



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> 1. What kind of obstructions would appear to represent $\omega$ in the form $\omega=\sum \overbrace{df\_i \otimes df\_i \otimes \ldots \otimes df\_i}^{k \;\text{times}}$ for some smooth functions $f\_1,f\_2,\ldots,f\_n$. When $k=2$ then this tensor is the pull back of the standard flat metric. Hence non vanishing of "Curvature" would be an obstruction for such a representation. What kind of tensors play the same role as curvature for $k>2$?
> 2. What kind of obstructions would appear to have a change of coordinate $\phi$ such that $\phi^\*{\omega}=\sum\_{i=1}^n g\_i \overbrace{dx\_i\otimes dx\_i\otimes \ldots \otimes dx\_i}^{k\;\text{times}}$ for some smooth functions $g\_1,g\_2,\ldots, g\_n$.
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A kind of "Curvature tensor" for higher dimensional tensors
Any $k$-linear symmetric map $\alpha:(\mathbb R^n)^k \to \mathbb R$ is the polarization of a homogeneous polynomial $\bar \alpha: x\mapsto \alpha(x,x,x\dots,x)$ of degree $k$ on $\mathbb R^n$.


First question: You ask whether any homogeneous polynomial of degree $k$ is of the form $\lambda\_1 x\_1^k + \lambda\_2 x\_2^k +\dots+ \lambda\_n x\_n^k$ after a suitable orthogonal coordinate change on $\mathbb R^n$. Obviously, the answer is no.


Any decomposition into real irreducible factors would survive any coordinate transformation, for example. So $\{\bar\alpha = 0\}$ can be a union of cones, whose number and their degrees are invariants. You might want to assume that the polynomial $\bar\alpha$ is irreducible over the reals, for example. Then look at the real algebraic geometry classification of irreducible real codimension one varieties of degree $k$ in $\mathbb R\mathbb P^{n-1}$.


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A point of view which might be more suited to your question: Determine first the orbit structure of the action of $GL(n)$ on the space $Pol^k[\mathbb R^n]$ of homogeneous polynomials of order $k$ on $\mathbb R^n$. If $k=1$ there is only 0 and $(\mathbb R^n)^\*\setminus \{0\}$. If $k=2$ rank and signature are full invariants and each orbit consisting of non-degenerate inner products is open. If $k>2$, in general, you do not have open orbits by dimension reasons. So you have uncountably many types of symmetric tensor fields.
Polynomial invariant theory does not help here since there are no polynomial $GL(n)$-invariants by Weyl's invariant tensor theorem. There are many $O(n)$-invariants, namely all polynomials in $g^{-1}(x\_i,x\_j)$, using the $O(n)$-invariant inner product $g$ on $\mathbb R^n$.


