Continuity at a point for function of two variables

Question

If a function of two variables is discontinuous at a particular point, say $(x,y)$, does this mean that the graph of that function has some hole around the point $(x,y,f(x,y))$? Is there any break in the graph at this point in certain direction?

This question arises because I have one function which is discontinuous at $(0,0)$ but all of its partial derivatives and directional derivatives exist at $(0,0)$. While calculating its partial or directional derivatives, we naturally look in a certain plane with that point and specified direction and calculate the slope of the tangent line (as you would with one variable). In my example I have all directional derivatives, which seems to imply that there is no break around $(0,0,f(0,0))$ in any direction. Then why is the function discontinuous at $(0,0)$?

Answer 1

Example 1 (one variable): Define $f(x) = \sin (1/x),x\ne 0,$ $f(0) =0.$ Then $f$ is (badly) discontinuous at $0.$ But there is no hole or break in the graph of $f.$ In fact the graph of $f$ is a connected subset of $\mathbb R^2.$

Example 2 (two variables): Define $f(x,x^2) = 1$ for real $x\ne 0.$ Define $f(x,y)=0$ everywhere else. Then all directional derivatives of $f$ at $(0,0)$ exist and are $0.$ But $f$ is discontinuous at $(0,0),$ as $\lim_{x\to 0} f(x,x) =0,$ while $\lim_{x\to 0} f(x,x^2) =1.$

Answer 2

Consider $f(x,y) = \frac{x^2+y^2}{x^2+y^2}$, using Cartesian coordinates. This function is undefined at $(0,0)$ so it cannot be continuous there. However, it is constantly $1$ everywhere else.

Now consider $g(x,y) = \frac{x^2-y^2}{x^2+y^2}$. What's happening at $(0,0)$?