Partial derivatives must exist and be continuous on all defined points of $f$ for $f$ to be differentiable? Today my professor explained that $f(x,y)=\frac{2xy}{(x^2+y^2)^2}$ is differentiable even though for $(x,y)=(0,0)$ $f$ is not defined.
The partial derivatives are $\frac{\partial f}{\partial x}=\frac{2y^3-6x^2y}{(x^2+y^2)^3} \text{ and } \frac{\partial f}{\partial y}=\frac{2x^3-6y^2x}{(x^2+y^2)^3}$. So, they exist.
Doesn't differentiability require the partial derivaties to be continuous? My professor said that we ignore $(0,0)$ on the partial derivatives because it isn't defined for $f$. I was hoping someone here could fully explain this.
Also could someone explain the difference between being in $C^1$ and just being differentiable?
 A: The first point to make is that we should care about the domains of functions. Usually, functions are defined simply by formulas, just as you wrote $f(x,y)=\dfrac{2xy}{(x^2+y^2)^2}$, and we understand its domain to be the largest in which the formula makes sense. In this example, the domain is $U=\mathbb{R}^2\setminus\left\{(0,0)\right\}$. This function cannot be extended to $\mathbb{R}^2$ to a continuous function. (Try to prove this: show that the limit $\lim_{t,s\rightarrow 0}f(s,t)$ does not exist.)
Now, forgetting some technicalities, the partial derivative $\dfrac{\partial f}{\partial x}$ is defined to be the function given by
$$\dfrac{\partial f}{\partial x}(x,y)=\displaystyle\lim_{t\rightarrow 0}\dfrac{f(x+t,y)-f(x,y)}{t},$$
as you know, and its domain is the set of points $(x,y)$ for which such limit exists, so $\dfrac{\partial f}{\partial x}:U\rightarrow\mathbb{R}$ is a function on its own right. In this case, $\dfrac{\partial f}{\partial x}(x,y)=\dfrac{2y^3-6x^2y}{(x^2+y^2)^3}$ for all $(x,y)\in U$.
The proper definition of a "differentiable function" may be a bit confusing for you, but let us say that a function is differentiable if all its partial derivatives exists (this is not the correct meaning). On the other hand, a function is said to be of class $C^1$ iff all its partial derivatives exist in the whole domain and they are all continuous functions.
From the very definition we already can see that $C^1$ functions are differentiable, but not all differentiable functions are $C^1$: For example, take $g:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $g(x,y)=x^2\sin(1/x)$ if $x\neq 0$ and $g(0,y)=0$. Then $g$ is differentiable and, for $(x,y)\in\mathbb{R}^2$,
$$\frac{\partial g}{\partial y}(x,y)=0;$$
$$\frac{\partial g}{\partial x}(x,y)=\begin{cases}2x\sin(1/x)-\cos(1/x)&\text{, if }x\neq 0\\0&\text{, if }x=0\end{cases}.$$
Notice that $\displaystyle\lim_{x\to 0}2x\sin(1/x)-\cos(1/x)$ does not exist, $\partial g/\partial x$ is discontinuous at the points $(0,y)$ ($y\in\mathbb{R}$) and $g$ is not $C^1$.
A: If a function is differentiable at a point $p$, then not only is it defined at $p$ it is also continuous at $p$. This is a classical result following almost immediately from the definition of differentiability. 
Functions in $C^1$ are those functions with a continuous first derivative. Not all differentiablefunctions have continuous first derivatives, so not every differentiable function is in $C^1$. 
