Do a pair of orthogonal directions with slopes equal to zero imply $\nabla f = 0$? The definition given on wikipedia for a saddle point is

is a point on the surface of the graph of a function where the slopes (derivatives) in orthogonal directions are both zero (a critical point), but which is not a local extremum of the function.

Looking at their picture (https://en.wikipedia.org/wiki/Saddle_point#/media/File:Saddle_point.svg) it's clear that $f_x(0,0) = 0$ and $f_y(0,0)=0$ .... just from taking a look. What isn't as clear is that $D_u f(0,0) = 0$ and that $D_v f(0,0)=0$ whenever $u\perp v$ (for unit $u$ and $v$). However if I turn the graph so that $\langle 1,0\rangle \mapsto u$ and $\langle 0,1\rangle \mapsto v$ then I would have $D_u f(0,0) = 0$ and that $D_v f(0,0)=0$ but then I don't see why $f_x(0,0)=0$ and $f_y(0,0)=0$ in that coordinate system.
Put another way.....  Is it true that $\nabla f(x_0,y_0)=0$ if and only if $D_u f(x_0,y_0)= D_v f(x_0,y_0)=0$ for some (and hence any) pair of unit vectors $u,v$ such that $u\perp v$? 
If this is true then in the boxed sentence above I could write instead "a point where $\nabla f=0$ but which ..."
 A: Remember than having an extrema of a differentiable function $f : \mathbb{R}^k \to \mathbb{R}$ at a point $p$ implies $\nabla_p f  = 0$. The converse however is not true, and so even if your question can be answered affirmatively, this says nothing about the former.
Now, as for your question, the answer is yes if $f$ is differentiable and $k = 2$ for the following reason: since $v \perp w$, the set $\{v,w\}$ is linearly independent and therefore a basis of $\mathbb{R}^2$. Since $f$ is differentiable, we have that for any point $p$ the directional derivative with direction $u$ is
$$
f_u(p) = \langle\nabla_p f, u \rangle \tag{1}
$$
Now, if $x$ is any vector, there exist $a,b \in \mathbb{R}$ so that $x =av+bw$ and thus,
$$
f_u(p) = \langle\nabla_p f, u \rangle = \langle\nabla_p f, av+bw \rangle = a\langle\nabla_p f,v\rangle+b\langle \nabla_p f, w \rangle = 0.
$$
In particular, the partial derivatives are zero, by taking $u = e_j$ with $1≤j≤2$. The other direction comes directly from $(1)$.
If $k>2$, the map 
$$
f(x_1, \dots, x_k) =  x_k
$$ 
has nonzero gradient but the partial derivatives $f_1$ and $f_2$ (which in particular are directional) are zero and $e_1 \perp e_2$.
