 Solution

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