Vtyjkyuk

Let $f colon mathbbR^2 to mathbbR$ be continuous differentiable and $x inmathbbR^2$ with $operatornamegrad f(x) neq 0$. Let $v,w in mathbbR^2$ be direction vectors.

If $D_v f(x) = D_w f(x)=0$ then are $v$ and $w$ linearly dependent.

For my proof I use the following theorem:

Let $U subset mathbbR^m$ be open and $f colon U to mathbbR$ continuous differentiable. Then it holds for $x in U$ and $v in mathbbR^m$ with $|v| = 1$ that $D_v f(x) = langle v, operatornamegrad f(x) rangle$.

Proof:

With the theorem above and the assumption, we know that
$$
langle v, operatornamegrad f(x) rangle
= langle w, operatornamegrad f(x) rangle.
$$
It holds that
beginalign*
langle v, operatornamegrad f(x) rangle
&oversetphantomD_vf(x)=D_wf(x)=
sum_i=1^2 v_i cdot fracpartial fpartial x_i
= v_1 cdot fracpartial fpartial x_1
+ v_2 cdot fracpartial fpartial x_2 \
&oversetD_vf(x)=D_wf(x)=
sum_i=1^2 w_i cdot fracpartial fpartial x_i
= w_1 cdot fracpartial fpartial x_1
+ w_2 cdot fracpartial fpartial x_2
= 0
endalign*
Therefore
beginalign*
&,
v_1 cdot fracpartial fpartial x_1
+ v_2 cdot fracpartial fpartial x_2
= w_1 cdot fracpartial fpartial x_1
+ w_2 cdot fracpartial fpartial x_2 \
iff&,
(v_1-w_1) fracpartial fpartial x_1
+ (v_2-w_2) fracpartial fpartial x_2
= 0
endalign*

Since $operatornamegrad f(x) = left(fracpartial fpartial x_1, fracpartial fpartial x_wright) neq 0$ this is a non trivial linear combination of $0$.

Hence $v$ and $w$ are linearly dependent.

Is this proof correct?
I am actually not sure if I used the correct scalar product $langle cdot, cdot rangle$.

Thanks in advance.

For given $x$ in the domain of $f$ the derivative $df(x)$ (also denoted by $f'(x)$, or similar) is a linear map acting as
$$df(x):quad mathbb R^2tomathbb R,qquad Xmapstolanglenabla f(x),Xrangle ,$$
whereby $langlecdot,cdotrangle$ denotes the standard scalar product in $mathbb R^2$, as in the question.
Since $nabla f(x)ne0$ by assumption this map has rank $1$; hence its kernel $K$ has dimension $2-1=1$. We are told that both $v$ and $w$ are in $K$; therefore $v$ and $w$ have to be linearly dependent.

Note that this argument does not work for an $f:>mathbb R^ntomathbb R$ with $n>2$.