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.
 A: $\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 $(0,0)$.
A: $f(x,y)$ isn't discontinuous at $(0,0)$.  How do I know?
First of all, can you get a picture of this function in your head?  I can.  Since $f: \Bbb R^{2} \to \Bbb R$, you should think of this function as first taking the $XY$-plane in $3$D space (let it be the horizontal axes), and warp that plane.  In this case, we are warping the plane like a taco.  If you are still confused, go to the WolframAlpha website and type "plot f(x,y) = y^2".  Our function looks almost like this, except when $y$ is irrational, we have the output $0$, so it's kind of like two planes -- one is a constant $0$, and the other is the taco, above the constant one.
Now, from the picture, it's clear that any path you take to $0$, the function approaches $0$, too, so the function has to be continuous at $(0,0)$.
Similarly, if you think about the picture, the partial derivative with respect to $x$ exists and is $0$, while the partial derivative with respect to $y$ exists and is $0$, because the limit $\lim \limits_{h \to 0} \dfrac{f(x, 0 + h) - f(x,0)}{h} = 0$.  
The only thing you need to worry about while checking this limit is the paths going to $0$. $\lim \limits_{h \to 0, h \in \Bbb Q} \dfrac{f(x, 0 + h) - f(x,0)}{h} = \lim \limits_{h \to 0, h \in \Bbb Q} \dfrac{h^{2} - 0}{h} = 0$, 
and
$\lim \limits_{h \to 0, h \not \in \Bbb Q} \dfrac{f(x, 0 + h) - f(x,0)}{h} = \lim \limits_{h \to 0, h \not \in \Bbb Q} \dfrac{0 - 0}{h} = 0$.
