Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives? Does there exist a function that is differentiable everywhere with everywhere discontinuous  partial derivatives?
I just had this doubt, talking about first order partials.

Yes it does. Take $$
f\left( {x,y} \right) = \left\{ \begin{array}{l}
 x^2 \sin \left( {\frac{1}{x}} \right) + y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \ne 0 \\ 
 x^2 \sin \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0,y = 0 \\ 
 y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y \ne 0 \\ 
 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y = 0 \\ 
 \end{array} \right..
$$ Then, $f$ is differentiable everywere however its partial derivatives are discontinuous 
$$f_x=
\left\{ \begin{array}{l}
 2x\sin \left( {\frac{1}{x}} \right) - \cos \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,x \ne 0 \\ 
 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \\ 
 \end{array} \right.
$$  and $$f_y=\left\{ \begin{array}{l}
 2y\sin \left( {\frac{1}{y}} \right) - \cos \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,y \ne 0 \\ 
 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 0 \\ 
 \end{array} \right.$$ which are discontinuous at the origin $(0,0)$. This example given in the book of B. Gelbaum and J. Olmsted, "Counterexample in Analysis", 3rd ed., page 119.
