# "Negation" test for differentiability of a multivariate function

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.
• Should the denominator of $\psi$ be $\lVert (x-x_0, y - y_0) \rVert$?