Coincidence of directional derivatives and derivatives along paths? $\newcommand{\R}{\mathbb{R}}$
$\newcommand{\al}{\alpha}$
Let $\Phi:\R^2 \to \R$ be a function which has partial derivatives at the point $(x_0,y_0)$. Let $\al:I \to \R^2$ be a path satisfying:
$\al(0)=(x_0,y_0),\dot \al(0)=v=(v_1,v_2)\neq (0,0)$.
Assume $\Phi \circ \al$ is differentiable at $t=0$ and the directional derivative  of $\Phi$ at direction $v$ ($d_v \Phi(x_0,y_0)$) exists.
Is it true that $(\Phi \circ \al)'(0)=d_v \Phi(x_0,y_0)$?
Update:
As noted by user251257, if $\Phi$ is Lipschitz then both derivatives agree. So, in order to find a counter-example (where they differ) we need to search for non-Lipschitz functions, which still has partial derivatives at the point $(x_0,y_0)$.
This is not trivial, since a function with bounded partial derivatives is Lipschitz (as proved here). So, our search is narrowed down to functions which do not have partial derivatives in a neighbourhood of $(x_0,y_0)$, or that there partial derivatives are unbounded in some neighbourhood of it.
Note:
In general, it can happen that either one of $(\Phi \circ \al)'(0),d_v \Phi(x_0,y_0)$ exists without the other.
Example 1: $d_v \Phi(x_0,y_0)$ exists ,$(\Phi \circ \al)'(0)$ does not.
(essentially taken from here)
$(x_0,y_0)=(0,0),
\Phi(x,y) =
\begin{cases}
 \frac{x^2y}{x^4 +y^2}, & \text{if $(x,y) \neq 0$}\\
0, & \text{if $(x,y) = (0,0)$} \\
\end{cases}, y(x)=x^2
$
$\Phi$ has directional derivatives at every direction at $(0,0)$, but for $\al(x)=(x,x^2)=(x,y(x))$ we get: $F(x)=\Phi \circ \al(x)=\begin{cases}
 \frac{1}{2}, & \text{if $(x,y) \neq 0$}\\
0, & \text{if $(x,y) = (0,0)$} \\
\end{cases}
$
 is not differentiable at $x_0=0$.
Example 2: $(\Phi \circ \al)'(0)$ exists ,$d_v \Phi(x_0,y_0)$ does not.
Update: this exmaple is wrong! I am not sure yet if a suitable example exists
Take $(x_0,y_0)=(0,0),\al(t)=(t,t^3+t),\Phi(x,y)=\sqrt{|xy|}$, $\Phi$ has partial derivatives ($\Phi_x(0,0)=\Phi_y(0,0)=0)$. $\Phi(\al(t))=\sqrt{t^4+t^2}$,so $(\Phi \circ \al)'(0)=1$. However, the directional derivative at direction $v=(1,1)$ does not exist.

If $\Phi$ is Lipschitz continuous on a neighborhood of $\alpha(0)$, then both derivatives do agree.
Updated counter example:
Consider following example: 
\begin{align}
\Phi(x,y) &= \sqrt[3]{xy}, \\  
\alpha(t) &= (t(1-t), t^2). 
\end{align}
Then, we have $\Phi(\alpha(t)) = t\sqrt[3]{1-t}$. On the other hand we have $\alpha'(0) = (1,0)$ and $\Phi(\alpha(0) + t\alpha'(0)) = \Phi(t, 0) = 0$. So both derivatives differ.
