Conditions for differentiability for many variables
For single variable functions we learn that being differentiable implies in being continuous. We also learn that being continuous doesn't imply in being differentiable because there are some exceptions to the rule. For multivariable functions differentiability also requires continuity, but we also have specific cases where we can calculate partial derivatives while the function is also discontinuous at a point.
[math]\displaystyle{ f(x) = |x| }[/math]. This single variable example shows that the limits exists at the origin, the function is continuous there and yet, we can't differentiate there because the rates of change to the left and to the right mismatch each other. Visually there is a sharp edge at the origin. [math]\displaystyle{ f(x,y) = \sqrt{x^2 + y^2} }[/math]. The same behaviour in two variables. The function is continuous at the origin, but the sharp edge means that we can't differentiate it there.
[math]\displaystyle{ f(x,y) = \begin{cases} 0 & \text{if} & (x,y) = (0,0) \\ \frac{xy}{x^2 + y^2} & \text{if} & (x,y) \neq (0,0) \end{cases} }[/math]
If we calculate [math]\displaystyle{ \frac{\partial f(0,0)}{\partial x} }[/math] and [math]\displaystyle{ \frac{\partial f(0,0)}{\partial y} }[/math] using the definition we do find valid results. But we also know, from calculating limits with different paths, that the function is discontinuous at the origin. With this example we have shown that the existence of partial derivatives doesn't guarantee that the function is continuous.
Differentiability
From a geometrical perspective, when a single variable function is differentiable it means that we can approximate it using a straight line in a small interval around a point. That's the idea of finding the tangent line. For two variables what is the equivalent to a line with one more dimension? It's a plane. Therefore, for a multivariable function to be differentiable it has to accept a linear approximation, a tangent plane around a point (beyond 3D the algebra exists, but we can't visualize the function). If it's rate of change in some direction is abrupt, much like a sharp edge, then we can't use a linear approximation there and the function is not differentiable at that point.
With concepts from linear algebra and analytical geometry we need a point and two linearly independent vectors to define a plane. For one variable we have this form:
[math]\displaystyle{ f(x) = f(x_0) + f'(x_0)(x - x_0) }[/math]
Where we have that [math]\displaystyle{ y = mx + b }[/math] is the equation of a straight line. The derivative being the angular coefficient. Extending it to a plane:
[math]\displaystyle{ z = f(x_0,y_0) + A(x - x_0) + B(y - y_0) }[/math]
Where A and B are, respectively, the angular coefficients in each direction. Notice that by thinking in analytical geometry, the above equation has a point of origin and two vectors.
For a single variable we use the linear approximation to prove that, when the difference between the linear approximation and the derivative itself at the same point is zero, then the function is continuous. By analogy, if the difference between a two variable function and its linear approximation is zero, the function is continuous. That is to write [math]\displaystyle{ f(x,y) - z = 0 }[/math], where [math]\displaystyle{ z }[/math] is the equation of the plane.
Now we need the directional derivative to show that if the function is differentiable, then it must be continuous. The partial derivatives won't suffice because they care about one direction parallel to the axis, but we are required to think on the derivative in any arbitrary direction, which is what the directional derivative is.
