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!
Best Answer
To complement user's answer, I would like to point out that the example in the OP is even more striking since not only do partial derivatives $\partial_1f(0,0)$ and $\partial_2f(0,0)$ exists, but also the directional derivative of the function $f$ at $\boldsymbol{0}=(0,0)$ along any direction $\mathbf{v}=(h,k)$ exists:
$$\partial_\mathbf{v}f(0,0):=\lim_{t\rightarrow0}\frac{f(\boldsymbol{0}+t\mathbf{v})-f(\boldsymbol{0})}{t}=\lim_{t\rightarrow0}\frac{1}{t}\frac{t^3k(h^2+k^2)}{t^2(h^2+k^2)}=k$$
So to add to other solutions:
A function $f$ may be
and yet not be differentiable.