Proof weird function is discontinuous/has no partial derivatives. I'm asked to analyze the continuity and existence of partial derivatives at the origin, and even though it seems pretty obvious that this function is discontinuous at that point, I can't seem to prove it via the standard limits. Any tips are welcome.
$$ f(x,y) =
  \begin{cases} 
      \hfill y^2    \hfill & y\in \Bbb Q \\
      \hfill 0 \hfill & y\in \Bbb R -\Bbb Q \\
  \end{cases}
$$
E: Could anyone graph this function with mathematica (or any software) I'd like to know how it looks, but I don't have mathematica at home and don't know how to use other software.

$\forall h\neq 0,\quad \left | \frac{f(0,0+h)-f(0,0)}{h} \right |<\left | \frac{h^{2}}{h} \right |=\vert h\vert $ so if we let $h\to 0$, we see that $f_y(0,0)=0$.
Likewise, 
$\forall h\neq 0,\quad \left | \frac{f(0+h,0)-f(0,0)}{h} \right |=\left | \frac{0}{h} \right |=0$ so $f_x(0,0)=0$.
Now, $\vert f(x,y)-f(0,0)\vert <y^{2}$ which is $<\epsilon $ as soon as $(x,y)\in  B_{\sqrt \epsilon }(0,0)$ so $f$ is continuous at $(0,0)$.
