If the directional derivative of the function $g$ exists in any direction If the directional derivative of a function $g$ exists for any $\boldsymbol{\theta}$ in any direction ${\bf d}$, and can be expressed as:
\begin{equation}
\lim_{\tau \rightarrow 0} \frac{g(\boldsymbol{\theta} + \tau {\bf d}) - g(\boldsymbol{\theta})}{\tau} =  {\bf \Delta}^\top {\bf d}  
\end{equation}
where ${\bf d}$ denotes any direction and $\tau$ is a scalar.
My question is: does this mean $g(\cdot)$ is differentiable at everywhere?
And does this imply $\nabla g(\boldsymbol{\theta}) = \Delta$?
My understanding is that for any differentiable function $f(\cdot)$,
its directional derivative $\nabla_{\bf d} f(\boldsymbol{\theta})$
can be expressed as $\nabla_{\bf d} f(\boldsymbol{\theta}) = \nabla^\top f(\boldsymbol{\theta}){\bf d}$,
where $\nabla f(\boldsymbol{\theta})$ is the gradient of $f$ evaluated at $\boldsymbol{\theta}$. The first equation means the directional derivative of $g(\boldsymbol{\theta})$ can be written in the same form of a differentiable function.
So does this implies $g(\boldsymbol{\theta})$ is differentiable at any $\boldsymbol{\theta}$,
and its gradient is $\nabla g(\boldsymbol{\theta}) = \Delta$?
I can not guess any function which is not differentiable but its directional derivative can be written as $\Delta^\top {\bf d}$, for a constant vector  $\Delta$  and  any vector
${\bf d}$.
Can anyone help me? Thank you in advance!

The function
$$(x,y)\mapsto\begin{cases}0&\text{if }|y|\ge x^2 \\ \left(\frac{y}{x^2}\left(1 - \frac{y}{x^2}\right)\right)^2&\text{if }|y| \le x^2 \end{cases}$$
is differentiable everywhere except at $\theta=(0,0)$ and at that point, it is discontinuous but has directional derivatives equal to $0$ in every direction ${\bf d}.$
The proof is the same as for this variant.
