For a two variable function to be differentiable, does its partial derivatives have to be continuous? I am currently doing an Analysis-Differential Forms module and one of the questions states:
"Give an informal interpretation of what it means for a function $f(x,y)$ to be differentiable at point $c=(c_1,c_2)$"
The tutor gave this answer:
Taking a neighbourhood about $f(c_1,c_2)$. The function is differentiable if $f_x$ and $f_y$ (partial derivatives) are continuous and exist for all $x$ and $y$ values in that neighbourhood.
My question is, does its partial derivatives have to be continuous so that $f(x,y)$ is differentiable?
 A: Your tutor gave a sufficient condition for differentiability, not a necessary one. The answer to your question is "no."
See, for example, this page, which gives
$$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$$
as a counterexample.

As for "give an informal interpretation," the answer should be something like: $f(x,y)$ is differentiable at $c$ if $f$ is well-approximated by a linear function near $c$. Or in other words, if the graph of $f$ has a well-defined tangent plane at $c$.
A: By definition, $f$ is differentiable at $c$ if there exist a (unique) linear map $L\colon \mathbb{R}^2 \longrightarrow \mathbb{R}$ such that
$$
f(c_1+h,c_2+k) = f(c_1,c_2) + L(h,k) + \rho(h,k)
$$
with $\lim\limits_{(h,k)\to (0,0)} \frac{\rho(h,k)}{\|(h,k)\|} =0$. An informal interpretation of this is that a function $f$ is differentiable at $c$ if it has a well defined linear approximation in a neighborhood of $c$.
One can prove that $L(h,k) = \frac{\partial f}{\partial x}(c) h + \frac{\partial f}{\partial y}(c) k$ 
Actually, the partial derivatives being continuous imply differentiability but not the converse. 
