How to understand the proof of 'continuous partial derivatives implies differentiability at $a$' 
Theorem: Let the partial derivatives of $f$ exists in a neighbourhood around $a$, and let the partial derivatives be continuous (ONLY NEEDS TO BE) at $a$. Then, $f$ is differentiable at $a$.

I understand the proof, yet I'm having a difficulty getting an intuition in the 3D case. Here is how I would describe the general ideas of the proof:

*

*There is $B(a,\delta)$ where the partial derivatives are defined, and $\forall i, |Df_i(x) - Df_i(a)|< \varepsilon$, if $|x-a|<\delta$.

*For any point $x$ inside the ball, we join $a$ to $x$ (assuming $\mathbb{R}^p \to \mathbb{R}$) $p$ line segments in a 'taxi cab' way, each segment is parallel to exactly one of the axes.

*Suppose points $u_1,u_2 \in \mathbb{R}^p$ are the endpoints of one of these line segments. Using MVT, in these line segment there exists a point $c_i \in \mathbb{R}^p$ such that $|f(u_1)-f(u_2) - Df_i(a)|  \leq \varepsilon |x-a|$. Here we used the continuity of the partial derivatives, and the fact that all of the line segments are inside of $B(a,\delta)$.

*Then we just triangle these inequalities for all co-ordinates, and we arrive to the definition of differentiability of $f$ at $a$.

The proof roughly goes like that. However, I'm having issues interpreting this geometrically when $f:\mathbb{R}^2\to\mathbb{R}$. Can someone give me an idea?
 A: Differentiability at $(a,b) \in \mathbb{R}^2$ roughly means that you can get a good approximation of $f(a+h,b+k)$ using the partial derivatives $f'_x(a,b)$ and $f'_y(a,b)$:
$$
f(a+h,b+k) - f(a,b) \approx f'_x(a,b) \, h + \color{blue}{f'_y(a,b)} \, k
.
$$
(Where “good” means that the error can be controlled in a precise way, but now we're just talking intuition, so let's not get into that.)
Now if $f'_x(a,b)$ exists, that's enough to guarantee that you can get a good approximation of $f(a+h,b)$:
$$
f(a+h,b)-f(a,b) \approx f'_x(a,b) \, h
.
$$
This is just single-variable calculus along the line $y=b$.
Similarly, starting from the point $(a+h,b)$ and using single-variable calculus along the line $x=a+h$, you can get a good approximation of $f(a+h,b+k)$ using the partial derivative $f'_y(a+h,b)$ (if it exists):
$$
f(a+h,b+k)-f(a+h,b) \approx f'_y(a+h,b) \, k
.
$$
Adding up the last two equalities, we get
$$
f(a+h,b+k) - f(a,b) \approx f'_x(a,b) \, h + \color{blue}{f'_y(a+h,b)} \, k
,
$$
and now comes the point: if $f'_y$ is continuous at $(a,b)$, then we won't make much of an additional error by replacing $f'_y(a+h,b)$ with $f'_y(a,b)$, and this gives us the first approximation above (the one associated with differentiability).
(Note that we actually don't need both partials to be continuous at $(a,b)$;
it's enough that one of them is. Or in general that $n-1$ of the $n$ partials are; see this answer.)
