The partial derivatives seem continuous to me, but the function is not differentiable For example in the following function:
$$f(x,y)=
\begin{cases}
\frac{x^3+y^3}{x^2+y^2},(x,y)\neq 0\\ 0,(x,y)=0\end{cases}$$
The partial derivatives in $(0,0)$
for $x$ and $y$ are $1$
therefore the derivative exists and is continuous at a point
but the function is not differentiable in $(0,0)$.
How is it possible?
 A: No, the partial derivatives are not continuous in $(0,0)$. For $(x,y) \ne (0,0)$, you have:
$$f_x(x,y)=\dfrac{x\left(x^3+3y^2x-2y^3\right)}{\left(x^2+y^2\right)^2}$$
$$f_y(x,y)=\dfrac{y\left(y^3+3x^2y-2x^3\right)}{\left(y^2+x^2\right)^2}$$
And, as you correctly noticed, $f_x(0,0)=f_y(0,0)=1$. But:
$$\nexists\lim_{(x,y)\to (0,0)}f_x(x,y)$$
$$\nexists\lim_{(x,y)\to (0,0)}f_y(x,y)$$
For instance, consider $f_x(0,y)$ and $f_x(x,0)$ and similar restrictions for $f_y$.
A: Define,
\begin{align*}
f:\mathbf{R}^{2}&\longrightarrow \mathbf{R},\\
(x,y)&\longmapsto \displaystyle\begin{cases}\frac{x^{3}+y^{3}}{x^{2}+y^{2}},\quad (x,y)\not=(0,0),\\0,\quad (x,y)=(0,0) \end{cases}
\end{align*}

*

*Since $$\lim_{h\to 0}\frac{f((0,0)+h(1,0))-f(0,0)}{h}=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=1$$
and
$$\lim_{h\to 0}\frac{f((0,0)+h(0,1))-f(0,0)}{h}=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=1$$
So, indeed the partial derivative of $f$ there exists at $(0,0)$.

*However the differentiability requires something else and that is to ensure the following condition
$$\color{blue}{\lim_{(x,y)\to (0,0)}\frac{f(x,y)-f(0,0)-f_{x}(0,0)(x-0)-f_{y}(0,0)(y-0)}{\sqrt{(x-0)^{2}+(y-0)^{2}}}=0}\quad (\star)$$
But in this case we have
\begin{align*}\frac{\frac{x^{3}+y^{3}}{x^{2}+y^{2}}-x-y}{\sqrt{x^{2}+y^{2}}}&=\frac{\frac{(r\cos \theta)^{3}+(r\sin \theta)^{3}}{r^{2}}-(r\cos \theta)-(r\sin \theta)}{\sqrt{r^{2}}},\quad \begin{cases}x=r\cos \theta,\\ y=r\sin \theta\end{cases} r\in \mathbf{R}_{+}^{*}, \theta \in [0,2\pi[,\\
&=\frac{\frac{r^{3}(\cos^{3}\theta+\sin^{3}\theta)}{r^{2}}+r(-\cos \theta-\sin \theta)}{r},\\
&=\frac{r}{r}\left(\cos^{3}\theta+\sin^{3}\theta-\cos\theta-\sin\theta\right),\\
&=\cos^{3}\theta+\sin^{3}\theta-\cos\theta-\sin\theta
\end{align*}
Hence, for a example with $\theta=\pi/3\in [0,2\pi[$ we have
$$\cos^{3}\frac{\pi}{3}+\sin^{3}\frac{\pi}{3}-\cos\frac{\pi}{3}-\sin\frac{\pi}{3}=-\frac{3}{8}-\frac{\sqrt{3}}{8}\not=0$$
Therefore $(\star)$  is not necessarily $0$, and so $f$ is not differentiable at $(0,0)$.

A: The partial derivatives are not continuous at $(0,0)$:
$f$ is not even differentiable at $(0,0)$, since it is homogeneous of degree $1$ but not linear.
