I have a question that is formulated as below.
Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a function which has continuous second partial derivatives such that $$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0. $$
Suppose that $u$ and $v$ is a pair of orthogonal unit vectors in $\mathbb{R}^2$. Prove that $$ D_u(D_vf) + D_v(D_uf) = 0 \quad (\text{OR} \quad \nabla_u(f) + \nabla_v(f) = 0), $$
where $D_uf$ and $D_vf$ ($\nabla_uf$ and $\nabla_vf$) denote the directional derivative of $f$ along $u$ and $v$ respectively.
Now as you can see below, the answer assumes that we replace $v_1$ and $v_2$ with $-u_2$ and $u_1$. It also says you can assume without loss of generality, but in my mind, since you are assuming a specific case of orthogonality, wouldn't that actually take away generality of your argument? I would also like to know if there are other ways to solve this without replacing the elements of $\vec{v}$ as proposed in the answer. Below is the answer.
Answer:
Let $u = (u_1, u_2)$ and $v = (v_1, v_2)$ be a pair of orthogonal unit vectors in $\mathbb{R}^2$. Without loss of generality, assume that $v$ can be obtained by rotating $u$ by $90^\circ$ in anticlockwise direction. Then, we have $u_1^2 + u_2^2 = 1$ and $v = (v_1, v_2) = (-u_2, u_1)$.
$$ D_vf = \nabla_vf = (u_1, u_2) \cdot \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = u_1 \frac{\partial f}{\partial x} + u_2 \frac{\partial f}{\partial y} $$
$$ D_u(D_vf) = \nabla_u(\nabla_vf) = (u_1, u_2) \cdot \left( \frac{\partial}{\partial x}\left(u_1 \frac{\partial f}{\partial x} + u_2 \frac{\partial f}{\partial y}\right), \frac{\partial}{\partial y}\left(u_1 \frac{\partial f}{\partial x} + u_2 \frac{\partial f}{\partial y}\right) \right) $$
$$ = (u_1, u_2) \cdot \left( u_1 \frac{\partial^2 f}{\partial x^2} + u_2 \frac{\partial^2 f}{\partial x \partial y}, u_1 \frac{\partial^2 f}{\partial x \partial y} + u_2 \frac{\partial^2 f}{\partial y^2} \right) $$
$$ = u_1^2 \frac{\partial^2 f}{\partial x^2} + 2u_1 u_2 \frac{\partial^2 f}{\partial x \partial y} + u_2^2 \frac{\partial^2 f}{\partial y^2} $$
Similarly, we have
$$ D_v(D_uf) = \nabla_v(\nabla_uf) = \left( -u_2 \frac{\partial}{\partial x} + u_1 \frac{\partial}{\partial y} \right)\left( u_1 \frac{\partial f}{\partial x} + u_2 \frac{\partial f}{\partial y} \right) $$
$$ = (-u_2)^2 \frac{\partial^2 f}{\partial x^2} + 2(-u_2)(u_1) \frac{\partial^2 f}{\partial x \partial y} + (u_1)^2 \frac{\partial^2 f}{\partial y^2} $$
$$ = u_2^2 \frac{\partial^2 f}{\partial x^2} - 2u_1 u_2 \frac{\partial^2 f}{\partial x \partial y} + u_1^2 \frac{\partial^2 f}{\partial y^2} $$
Then,
$$ D_u(D_vf) + D_v(D_uf) = \nabla_u(\nabla_vf) + \nabla_v(\nabla_uf) $$
$$ = \left( u_1^2 \frac{\partial^2 f}{\partial x^2} + 2u_1 u_2 \frac{\partial^2 f}{\partial x \partial y} + u_2^2 \frac{\partial^2 f}{\partial y^2} \right) + \left( u_2^2 \frac{\partial^2 f}{\partial x^2} - 2u_1 u_2 \frac{\partial^2 f}{\partial x \partial y} + u_1^2 \frac{\partial^2 f}{\partial y^2} \right) $$
$$ = (u_1^2 + u_2^2) \frac{\partial^2 f}{\partial x^2} + (u_2^2 + u_1^2) \frac{\partial^2 f}{\partial y^2} $$
$$ = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} $$
$$ = 0 $$
I guess you could argue that $u$ has only two perpendicular vectors (when controlling for the norm, but it does not really matter here) $v$ and $-v$ but since $D_{-v}$ = $-D_v$ (as linear operators), your proof will just add a - to both sides of the sum.