One of the "acceptance test" for differentiability of $f(x,y)$ at $(x_0,y_0)$ is to verify that its partial derivatives exists and are continuous in a neighbourhood. I wonder what are some of the ways to quickly negate differentiability of a function where all directional derivatives exist.
The following is obvious:
Given a function $z=f(x,y)$, we want to test its differentiability at $(x_0,y_0)$. Find the "candidate tangential plane" with normal vector $\vec{n}=(f_x(x_0,y_0),f_y(x_0,y_0),-1)$. Choose a path $(x,y(x))$ and evaluate tangent vector of the curve $(x,y(x),f(x,y(x)))$ at $(x_0, y_0)$ and check if it is normal to the normal vector $\vec{n}$.
Are there any simpler procedures?
Let $a=\frac{\partial f}{\partial x}(x_0,y_0)$ and $b=\frac{\partial f}{\partial y}(x_0,y_0)$. Consider the quotient$$\psi(x,y)=\frac{f(x,y)-f(x_0,y_0)-a(x-x_0)-b(y-y_0)}{\bigl\lVert(x-x_0,y-y_0)\bigr\rVert}.$$If there are two continuous paths $\gamma_1$ and $\gamma_2$ such that $\gamma_1(0)=\gamma_2(0)=(x_0,y_0)$ the limits$$\lim_{t\to0}\psi\bigl(\gamma_1(t)\bigr)\text{ and }\lim_{t\to0}\psi\bigl(\gamma_1(t)\bigr)$$are distinct (or if at least one of them doesn't exist), then $f$ is not differentiable at $(x_0,y_0)$.
Note this only implies that $f$ is not differentiable at that point; it is not equivalent to the assertion that $f$ is not differentiable there.