Differentiability — The Definition Unpacked
Recall: The Single-Variable Ancestor
In one-variable calculus the derivative $f'(a)$ is the number that makes the increment split cleanly:
$$ \Delta y = f(a+\Delta x)-f(a) = f'(a)\,\Delta x + \varepsilon\,\Delta x, \qquad \varepsilon \to 0 \text{ as } \Delta x \to 0. $$The linear part $f'(a)\,\Delta x$ is the best linear approximation; the remainder $\varepsilon\,\Delta x$ vanishes faster than $|\Delta x|$ itself. This is the template we now upgrade to two variables.
Definition — Differentiability at $(x_0,y_0)$
Let $z = f(x,y)$ be defined on an open set containing $(x_0,y_0)$, and suppose the partial derivatives $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ exist. Write the full increment as
$$ \Delta z = f(x_0+\Delta x,\, y_0+\Delta y) - f(x_0,y_0). $$We say $f$ is differentiable at $(x_0,y_0)$ if
$$ \Delta z = \underbrace{f_x(x_0,y_0)\,\Delta x + f_y(x_0,y_0)\,\Delta y}_{\text{linear part}} \;+\; \underbrace{E_x\,\Delta x + E_y\,\Delta y}_{\text{remainder}}, $$where $E_x \to 0$ and $E_y \to 0$ as $(\Delta x,\Delta y)\to(0,0)$.
What plays the role of the derivative? In one variable the derivative is a single number $f'(a)$. Here the pair $(f_x, f_y)$ evaluated at the point plays that role — these are the coefficients of the linear part, and they are uniquely determined by the differentiability condition.
What does "smallness of the remainder" demand? The error terms $E_x$ and $E_y$ must each tend to zero as $\|(\Delta x,\Delta y)\| \to 0$. Equivalently, the remainder $R = \Delta z - f_x\,\Delta x - f_y\,\Delta y$ satisfies
$$ \frac{|R|}{\sqrt{(\Delta x)^2+(\Delta y)^2}} \;\longrightarrow\; 0 \quad \text{as } (\Delta x,\Delta y)\to(0,0). $$
The remainder vanishes