Suppose that $u(x, y)$ has a local maximum at $(0,0) .$ Show that for any direction $\frac{\partial u(0,0)}{\partial \vec{V}}=0$.
So I want to show that directional derivative is zero. So finally I get ${u(ah,bh) - u(0,0)}$ / $h$ as limit $h$ tends to zero ; how to proceed after that.
The map$$\begin{array}{rccc}\varphi\colon&\mathbb R&\longrightarrow&\mathbb R\\&h&\mapsto&u(ah,bh)\end{array}$$has a local maximum at $0$, since $\varphi(0)=u(0,0)$ and $u$ has a local maximum at $(0,0)$. On the other and, $\varphi$ is differentiable at $0$, since$$\varphi'(0)=\lim_{h\to0}\frac{u(ah,bh)-u(0,0)}h,\tag1$$which is the directional derivative of $\varphi$ at $(0,0)$, with respect to the direction given by $(a,b)$. But then, since $\varphi$ has a local maximum at $0$, $(1)$ is equal to $0$. In other words, the directional derivative of $\varphi$ at $(0,0)$, with respect to the direction given by $(a,b)$, is equal to $0$.