= Solution
For $\phi=x^2+2y^2$, the <gradient> is $\nabla\phi=(2x,4y)$, and
$$
(2y,-x)\cdot(2x,4y)=0.
$$
Away from the origin, both $\nabla\phi$ and $\nabla u$ lie in the one-dimensional subspace perpendicular to $(2y,-x)$, so they are parallel. More precisely,
$$
\boxed{\nabla u=\lambda(x,y)\nabla\phi}
$$
for some scalar $\lambda$, including $\lambda=0$. If $\nabla u\ne0$, this can also be inverted to express $\nabla\phi$ as a multiple of $\nabla u$. At a <stationary point> of $u$ with $\nabla\phi\ne0$, that literal inverse wording is not valid; the geometric assertion is parallelism of the nonzero <gradient> directions.
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