Can a function have partial derivatives, be continuous but not be differentiable? 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!
 A: By differentiability theorem if partial derivatives exist and are continuos in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.
The existence of partial derivatives doesn't suffice.
In this case we can proceed by definition
$$ \lim_{(h,k) \to (0,0)} \frac{\dfrac{2h^2k+k^3}{h^2+k^2}-k}{\sqrt{h^2+k^2}}= \lim_{(h,k) \to (0,0)} \frac{2h^2k+k^3-kh^2-k^3}{\sqrt{(h^2+k^2)^3}}= \lim_{(h,k) \to (0,0)} \frac{h^2k}{\sqrt{(h^2+k^2)^3}}$$
which doesn't exist and therefore the function is not differentiable at $(0,0)$ and indeed partial derivatives are not continuous at that point.
A: 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

*

*continuous at some point $\mathbf{c}$,

*have (finite) directional derivatives at $\mathbf{c}$ along any vector $\mathbf{v}$ ($\partial_1f(\mathbf{c})$ and $\partial_2f(\mathbf{c})$ correspond to $\mathbf{v}=(1,0)$ and $\mathbf{v}=(0,1)$ respectively)

and yet not be differentiable.
