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!!

Continuty at (3,0) : When  $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$
Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at $(3,0)$ wrt $y$ and the partial derivative is $0.$
