I posted this problem earlier today, but now I have some other questions regarding the problem.
$$F(x,y)=\begin{cases} f(x,y) & (x,y)\neq (0,0) \\ 0 & (x,y)=0 \end{cases}$$
where $f(x,y)=\frac{x^2y}{x^2+y^2}$
*1.) Determine if f is continuous in origin.*
The function f(x,y) is not continuous in origin, because it is not defined in origin.
2.) Decide $F_1(x,y)$ and $F_2(x,y)$.
$$F_1(x,y)=\frac{2xy(x^2+y^2)-x^2y(2x)}{(x^2+y^2)^2}=\frac{2xy^3}{(x^2+y^2)^2}$$ $$F_2(x,y)=\frac{x^2(x^2-y^2)}{(x^2+y^2)^2} $$
But this derivative will only apply to $(x,y)\neq(0,0)$. Do I also have to find another expression for $F_1$ and $F_2$ when $(x,y)=(0,0)$, or should that be done in task number 3 ?
3. Determine the partial derivatives in origin and decide if $F$ is differentiable in origin.
So I'm a bit confused. I thought about that finding the partial derivatives in origin by using
$$ F_1(0,0)=\lim_{h\to0}\frac{F(0+h,0)-F(0,0)}{h}=\lim_{h\to 0}\frac{\frac{h^2*0}{h^2+0}}{h}=0 $$
$$F_2(0,0)=\lim_{h\to0}\frac{F(0,0+h)-F(0,0)}{h}=\lim_{h\to 0}\frac{\frac{0^2*h}{h^2+0}}{h}=0$$
I don't know if this is right, but how do I then determine if $F$ is differentiable in origin?
Your computation of the partial derivatives at $(0,0)$ is fine (and elsewhere too). Since they are both equal to $0$, if $F$ was differentiable at the origin, $F'(0,0)$ would be the null function. In other words, we would have$$\lim_{(x,y)\to(0,0)}\frac{F(x,y)}{\sqrt{x^2+y^2}}=0,$$which is equivalent to$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But this is not true, because if $y=x>0$,$$\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac{x^3}{2\sqrt2\,x^3}=\frac1{2\sqrt2}\neq0.$$