Partial derivatives exist and derivative does not Wikipedia states that partial derivatives of this function exist, while derivative doesn't. I understand that the problem lies at $x=-1$.
$$f(x,y) = \begin{cases}x & \text{if }y \ne x^2 \\ 0 & \text{if }y = x^2\end{cases}$$
The partial derivatives for this function are
$$f(x,y)'_x = \begin{cases}1 & \text{if }y \ne x^2 \\ 0 & \text{if }y = x^2\end{cases}$$
$$f(x,y)'_y = \begin{cases}0 & \text{if }y \ne x^2 \\ 0 & \text{if }y = x^2\end{cases}$$
Does the problem with the derivative arise because of the jump at $x = -1$?
EDIT:
Wikipedia link see section "Differentiability in higher dimensions"
 A: What is claimed in the Wikipedia article is that $f$ is has partial derivatives at the origin but is non-differentiable there, so this has nothing whatsoever to do with $x=-1$.
Indeed, $f(x,0)=x$ for all $x$ (including $x=0$), so $f'_x(x,0)=\frac{d}{dx}x=1$ for all $x$, in particular $$A=f'_x(0,0)=1.$$
And $f(0,y)=0$ for all $y$ (including $y=0$), so $f'_y(0,y)=\frac{d}{dy}0=0$ for all $y$, in particular $$B=f'_y(0,0)=0.$$
So the partials exist at the origin. (And it can be shown similarly that all directional derivatives exist, as they also claim.)
But $f$ is not differentiable at the origin, since the function
$$
\frac{f(x,y)-f(0,0)-Ax-By}{\sqrt{x^2+y^2}}
= \frac{f(x,y)-x}{\sqrt{x^2+y^2}}
= \begin{cases}
0 ,& y\neq x^2 \\
\dfrac{-x}{\sqrt{x^2+y^2}} ,& y=x^2
\end{cases}
$$
doesn't tend to zero as $(x,y)\to(0,0)$. (Along the curve $y=x^2$, $x>0$, it tends to $-1$.)
P.S. Your expression for the partials $f'_x$ and $f'_y$ are not correct; they don't exist at the points on the curve $y=x^2$, except at the origin.
