Let $f(x,y):R^2\rightarrow R$ to be the function such that $f(x,y)=\frac{x*y}{x^2+y^2}$?
Why of $f_1, f_2$ exists everywhere for $\frac{x*y}{x^2+y^2}$?
For example $f_1=f_x=\frac {\partial f(x,y)}{x}=\frac{y(x^2+y^2)-xy(2*x)}{(x^2+y^2)^2}$. When I consider the point where $(x,y)=(0,0)$, I didn't see why $f_1$ still exists.
You have to answer a question like this by returning to the original limit definition of the partial derivative. In other words, you get this result by returning to the first principle definition.
In this case you know that If you think about the definition of the derivative it's a bit clearer. Firstly the function is not well defined. I assume it should say: \begin{align*} f(x,y) = \begin{cases} 0 & (x,y)=(0,0)\\ \frac{xy}{x^2+y^2} & \mathrm{otherwise}\\ \end{cases} \end{align*} Now by applying the definition of partial differentiation, \begin{align*} f_1(0,0) = \frac{\partial f(x, y)}{\partial x}\Big|_{(0,0)} = \lim_{h\rightarrow0}\frac{f(0+h, 0) - f(0, 0)}{h} = \lim_{h\rightarrow0}\frac{\frac{0h}{h^2+0}-0}{h} = \lim_{h\rightarrow0}\frac{0}{h^3} = 0 \end{align*}
Similarly you can show that $f_2$ exists.
