= Solution
An <isotropic tensor> has unchanged components under every rotation of the Cartesian axes. Under proper rotations in three dimensions, rank one has \b[no nonzero isotropic tensor], rank two has $\boxed{T_{ij}=A\delta_{ij}}$, and rank three has $\boxed{T_{ijk}=B\epsilon_{ijk}}$. Here $\delta$ is the <Kronecker delta> and $\epsilon$ the <Levi-Civita symbol>. If invariance under reflections is also required for a true tensor, the rank-three case vanishes; $\epsilon$ is then a pseudotensor.
The <velocity gradient tensor> for the given field is
$$
D_{ik}=\frac{\partial u_i}{\partial x_k}=\begin{pmatrix}x_2^2&2x_1x_2&0\\0&x_3^2&2x_2x_3\\2x_1x_3&0&x_1^2\end{pmatrix}.
$$
Its symmetric part, the <strain-rate tensor> if $\mathbf u$ is interpreted as a velocity, is
$$
\boxed{e=\frac{D+D^T}{2}=\begin{pmatrix}x_2^2&x_1x_2&x_1x_3\\x_1x_2&x_3^2&x_2x_3\\x_1x_3&x_2x_3&x_1^2\end{pmatrix}.}
$$
The antisymmetric part has the specified ordering $D_{ik}-e_{ik}=\epsilon_{ijk}\omega_j$, with
$$
\boxed{\boldsymbol\omega=\tfrac12\nabla\times\mathbf u=(-x_2x_3,-x_1x_3,-x_1x_2).}
$$
The index ordering matters for this sign; it agrees with a local rotation velocity $\boldsymbol\omega\times\mathbf x$. This is the axial part of the <scalar, symmetric-traceless and axial tensor decomposition>, retaining the trace inside the symmetric part here.
At the specified point, $e=\begin{pmatrix}9&6&0\\6&0&0\\0&0&4\end{pmatrix}$. The upper block has characteristic polynomial $\lambda^2-9\lambda-36=(\lambda-12)(\lambda+3)$. Thus the principal values and unit <eigenvectors> are
$$
\boxed{12:\ \frac{(2,1,0)}{\sqrt5};\qquad -3:\ \frac{(1,-2,0)}{\sqrt5};\qquad 4:\ (0,0,1).}
$$
These mutually orthogonal directions are the <principal strain rates>' axes for this local field; reversing an eigenvector does not change its axis.
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