lim of f(x,y) for which const does it exist? Giving the following function:
when y!=0
$$f(x,y) = \frac{x^4+y^4}{y(x^2+y^2)} $$
else:
$$f(x,y)=A$$
I am asked for which value of A (costs) the following lim does exist.
$$\lim_{ x\to 0, y\to 0} f(x,y) = L$$ 
So I drew the function using geogebra 3d and I can easily see if A wasn't 0 then there is always some points to close to (0,0) which their values are far from the function which means that the lim doesn't exist.
But, my answer is wrong.
The graph of the function:

 A: This was bugging me for like the last half hour, but I'm pretty sure I found your mistake. The answer is that there is no value of $A$ for which the limit in question exists. 
You've already ruled out all values of $A$ other than $0$, although your logic of there are always some points to close to $(0,0)$ for which the value is far from $0$ isn't quite sound.
A better way to phrase it would be if there is a limit, it must be equal to $A$ (because paths along $y=0$ will have limit $A$ as $x$ approaches $0$), AND it must be equal to $0$ (set $x=y$ and take the limit).
So can we have a limit of zero when $A=0$? No.
Consider fixing the normal expression equal to $1$ (so taking a horizontal slice through the graph). Via wolfram alpha, it would look like this:

Which does intersect $(0,0)$, so that set of $x,y$ values represent a path to $(0,0)$ for which the limit of $f(x,y)\ne0$!  
In fact, we can show that $\lim \limits_{x\to0, y\to0} f(x,y)$ does not exist without even considering the value of $A$: We have a path ($x=y$) along which the limit is $0$, and a path ($1=f(x,y)$) along which the limit is $1$.
A: Along the path $y=x$ is the limit 
$$\lim\limits_{x\to 0}\frac{2x^4}{2x^3}=0,$$
while along the path $y=x^2\;$ we have
$$\lim\limits_{y\to 0}\frac{y^2+y^4}{y^2+y^3}=\lim\limits_{y\to 0}\frac{1+y^2}{1+y}=1,$$ therefore the limit does not exist. There is no suitable value of $A.$
