Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ have continious partial derivatives in a neighborhood of $\partial B(0,1)$. Show there exist at least 2 unit vectors such that for all $i,j$
$$u_i\frac{\partial f}{\partial x_j}(\vec{u}) = u_j\frac{\partial f}{\partial x_i}(\vec{u})$$
I thought using Lagrange Multipliers would be a good idea, to find an exteremum of a function with the constraint of a unit ball and some other manifold, which yields this equation. I'm not sure which function (I guess it would be $f$ itself) and constraint to use though.
I will give the answer in the lower dimensions $n=2,3$.
For $n=2$, take $$(u_1,u_2)=\pm \frac{1}{\sqrt{f^2_x+f^2_y}}(f_x,f_y).$$ This is a special case of the dimension $n=3$:
Since $$u_{i}\frac{\partial f}{\partial x_j}= u_{j}\frac{\partial f}{\partial x_i}$$ then, evidently, $$\nabla f\times u=0.$$ This means that $u$ is parallel to $\nabla f$. Actually the vector product of two vectors vanishes if and only if the vectors are parallel. So, take $$(u_1,u_2,u_3)=\pm \frac{\nabla f}{|\nabla f|}.$$