I have a function:
$$ f(x,y)= \begin{cases} \dfrac{2x^2y+y^3}{x^2+y^2} & \text{if $(x,y) \neq (0,0)$}\\ 0 & \text{if $(x,y) = (0,0)$}\\ \end{cases} $$
which I think I managed to show:
a) continuity at $(0,0)$
by $\lim_{(x,y) \to (0,0)} f(x,y) = 0$
b) has partial derivatives at $(0,0)$
by the definition of derivatives and found $f'_x(0,0) = 0, f'_y(0,0) =1$. Still not 100% sure if did this correctly.
c) not differentiable at $(0,0)$
by definition of differietable functions and that a limit didn't exist.
However, I feel like because of this I can tell more about the function. I'd like it if someone can confirm this. I assumed, that because it wasn't differentiable, the partial derivatives might not be continuous around $(0,0)$. $$\frac{\partial f}{\partial x} = \frac{2y^3x}{\left(x^2+y^2\right)^2}$$ $$\frac{\partial f}{\partial y} = \frac{y^4+y^2x^2+2x^4}{\left(x^2+y^2\right)^2}$$
Is that the case? I checked the limits $$\lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial x} \quad \text{and} \quad \lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial y}$$ and they don't seem to exist. What would happen if one existed but not the other? Is this possible? What would happen if the limit was something else than $0$ and $1$ I calculated in b)? Just not being continuous? I am just worried if the function really has partial derivatives in $(0,0)$.
Thank you in advance!