# Determining differentiability

It is clear that the function $\psi(x,y)=yx^{-1}$ is differentiable on $R^2-{0}$. But what about for $x=0$ -- what if I set $\psi(0,y)\equiv0$? Then how can I determine whether $\psi$ is differentiable there? Thanks.

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The new function is not continuous at any point of the form $(0,a)$; so it cannot be differentiable.
Approaching $(0,a)$, $a\neq 0$, along $x=0$ gives a limit of $0$. Approaching along any other line, $y = kx + a$ as $x\to 0$, gives \begin{align*} \lim_{x\to 0^+}\psi(x,kx+a) &= \lim_{x\to 0^+}\frac{kx+a}{x} = \left\{\begin{array}{ll} \infty & \text{if }a\gt 0\\ -\infty & \text{if }a\lt 0 \end{array}\right.\\ \lim_{x\to 0^-}\psi(x,kx+a) &= \lim_{x\to 0^-}\frac{kx+a}{x} = \left\{\begin{array}{ll} -\infty & \text{if }a\gt 0\\ +\infty & \text{if }a\lt 0 \end{array}\right. \end{align*}
At $(0,0)$, approaching along $x=ky$ gives a limit of $k$, so again the limit does not exist.
Since the function is not continuous at any point $(0,a)$, it's not differentiable either.