Continuity, partial derivatives and total differential 
Is it true that if I have some function $f(x,y)$, for which partial
derivatives exist but it's not continuous, that this function cannot be
differentiable (total differential doesn't exist)? But if my function
is continuous and the partial derivatives do exist (and are also continuous), then it must be
differentiable?

I am confused, for a long time. I have found some examples but I just want to be clear.
 A: Take the function
$$f (x,y)=(x^2+y^2)\cos (\frac {1}{\sqrt {x^2+y^2}}) $$
and $f (0,0)=0$.
it is differentiable at $(0,0) $ but the partial derivatives are not continuous at $(0,0) $.
A: Here is a rough sketch of how you prove continuous first partials implies differentiability. 
A function $f$ is differentiable at $x$ iff the change $f(x+h)-f(x)$ can be approximated by a linear map with a remainder term which is $o(h)$.
If a function has first partial derivatives defined at $x$ then it has arbitrary  directional derivatives at $x$
A generalisation of the mean value theorem shows that in a small region, you can approximate the change by $f(x+h) - f(x)$ about $x$ by the values of the directional derivative of $f$ in the direction $h$ on the line segment joining $x$ and $x+h$. 
If you construct a map $L$ which takes a vector and returns the directional derivative of $f$ in that direction, you will find that it is linear. If the partials are continuous, then you can prove that the remainder $f(x+h) - f(x) - Lh$ is $o(h)$ so $L$ is the derivative of $f$ at $x$.
I used this as a reference: https://www.math.ucdavis.edu/~hunter/book/ch13.pdf
A: In two dimensions, differentiability always implies continuity. Because of this, the idea of a multi variable function having existing partials without continuity seems impossible. However, there is a small workaround. If we consider the definition of a partial derivative (assuming 3-D for now) it looks like this:
$$\frac{\delta f(x, y)}{\delta x}=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$$
This can evaluate without issue without the function being "continuous with respect to $y$", or more precisely, without $f(k, y)$ being continuous for some $k$. However, the function must be "continuous with respect to $x$", or more precisely $f(x, k)$ must be continuous. An example of such a function can be found below:
$$f(x, y) =
\begin{cases}
x+1, &y\geq 0 \\
x^2, &y<0
\end{cases}
$$
Take for example point $(0, 0)$. The function evaluates to $1$ at this point, however it is not continuous since
$$\lim_{y\to 0}{f(0,y)}=0$$
However, if we plug this equation into the definition of the derivative, we get
$$\frac{\delta f(x, y)}{\delta x}=
\begin{cases}
1, & y\geq 0 \\
2x, & y<0
\end{cases}$$
In the end, you are right in asserting that this entire function cannot have all partial derivatives defined everywhere, as this would require continuity everywhere.

For your second question, this is true. The differentiability of a multi variable function is implied by the existence and continuity of all partial derivatives. Because of this, your function having existing continuous partial derivatives everywhere necessarily means it is differentiable.
