If $f(x,y)=9-x^2-y^2$ if $x^2+y^2\leq9$ and $f(x,y)=0$ if $x^2+y^2>9$ study what happens at $(3,0)$ 
If$$f(x,y)=\begin{cases}9-x^2-y^2&\text{if }x^2+y^2\leq9\\0&\text{if }x^2+y^2>9\end{cases}$$study the continuity and existence of partial derivative with respect to $y$ at point $(3,0)$.

The graph of the domain of $f$ is:

Continuity study at $(3,0)$:
$f(3,0)=9-3^2-0=0$, but I do not know how to find $$\lim_{(x,y)\to(3,0)}f(x,y).$$ I tried the following: $$\lim_{(x,y)\to(3,0)}f(x,y)=\left\{\begin{array}{l}\displaystyle\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_1\colon x^2+y^2=9}{\lim_{(x,y)\to(3,0)}}(9-(x^2+y^2))=9-9=0\\\displaystyle\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}f(x,y)=\underset{C_2\colon x^2+y^2\neq9}{\lim_{(x,y)\to(3,0)}}\text{??}=\text{??????}\end{array}\right.$$ but then I realized that the "curve" $C_2$ is actually NOT a curve but a set of infinite points, as shown in the previous image.
Existence of partial derivative with respect to $y$ at $(3,0)$:
I know that I need to study whether $$\frac\partial{\partial y}f(3,0)=f'((3,0);(0,1))=f_y(3,0)=\lim_{h\to0}\frac{f((3,0)+h(0,1))-f(3,0)}{h}=\lim_{h\to0}\frac{f(3,h)}{h}$$ exists or not, but I am not able to even find that limit.
Any help? Thanks!!

We have the piece-wise defined surface.  The two surfaces are continuous and if the limits are equal at the join, then the surface is continuous.
And since the two surfaces are continuous we can just plug $(3,0)$ in for each.
$\lim_\limits{(x,y)\to (3,0)} 9-x^2-y^2 = 0$
$\lim_\limits{(x,y)\to (3,0)} 0 = 0$
Another way to do it would be to say:
$\forall \epsilon >0, \exists \delta > 0 : d((x,y),(3,0))< \delta \implies |f(x,y)|<\epsilon$
$d((x,y),(3,0))$ is the distance between $(x,y)$ and $(3,0)$
And what distance metric do we want to use?  How about the taxicab metric?
Then the maximal distance some point $x_1,y_1$ can be from the point $(3,0)$ would be $(3+\pm \delta, \pm \delta)$
$f(x,y) < |6\delta| + |2\delta^2|$
Lets demand that $\delta < 1$  then $\delta^2 < \delta$
$\delta = \min (\frac {\epsilon}{8},1)$
