If $E$ is a topological space and $f:E\to\mathbb R$, say that $x_0\in E$ is a local minimum of of $f$ if $$f(x_0)\le f(x)\;\;\;\text{for all }x\in N\tag1$$ for some neighborhood $N$ of $x_0$. If $K\subseteq E$, say that $x_0\in E$ is a local minimum of of $f$ constrained on $K$ if $x_0$ is a local minimum of $\left.f\right|_K$ (where $K$ is equipped with the subspace topology).
Assume that $E$ is a normed $\mathbb R$-vector space, $\Omega\subseteq E$ is open and $f\in C^1(\Omega)$. It's easy to show that if $x_0\in\Omega$ is a local minimum of $f$, then $${\rm D}f(x_0)=0.\tag2$$
Now let $K\subseteq\Omega$ be a closed subspace of $E$ and $x_0\in\Omega$ be a local extremum of $f$ constrained on $K$. Can we conclude that $${\rm D}f(x_0)y=0\tag3$$ for all $y\in K$?
It should easily follow from $(2)$. In fact, $K$ equipped with the norm inherited from $E$ is a normed $\mathbb R$-vector space, $K$ is open in $K$, $\left.f\right|_K\in C^1(K)$ and $x_0$ is a local minimum of $\left.f\right|_K$. So, by $(2)$, $${\rm D}\left.f\right|_K(x_0)=0\tag3.$$
The claim would now follow if we could show that $${\rm D}\left.f\right|_K(x_0)y={\rm D}f(x_0)y\;\;\;\text{for all }y\in K\tag4.$$ Can we do this?
By Definition, $$\lim_{h \to 0}\frac{|f(x_0+h) - f(x_0) - [Df(x_0)]h|}{\|h\|} = 0$$ where $h$ varies over all of $E$. And by definition, $$\lim_{h \to 0}\frac{|f(x_0+h) - f(x_0) - [Df|_K(x_0)]h|}{\|h\|} = 0$$
Where now $h$ varies over $K$ only. When $h \in K$, the two formulas are the same, and therefore $Df|_K(x_0) = Df(x_0)|_K$, the restriction of $Df(x_0)$ to $K$.
Since $y \in K$, $$Df|_K(x_0)y = Df(x_0)|_Ky = Df(x_0)y$$