= Solution
For the <differentiability from continuous partial derivatives> argument, choose a small box around zero inside the given open set. For $h=(h_1,\ldots,h_p)$ in that box, telescope $f(h)-f(0)$ by changing one coordinate at a time. The one-dimensional <mean value theorem> on the $j$th coordinate segment gives
$$
f(h)-f(0)=\sum_{j=1}^p h_j f_j(\xi_j),
$$
where $\xi_j$ lies on that segment, and therefore $\|\xi_j\|_2\le\|h\|_2$. A zero-length segment contributes zero and needs no choice of intermediate point. Set $L(h)=\sum_j f_j(0)h_j$. For every $\varepsilon>0$, continuity of the finitely many <partial derivatives> at zero gives $|f_j(\xi_j)-f_j(0)|<\varepsilon/\sqrt p$ when $h$ is sufficiently small. Then the <Cauchy-Schwarz inequality> gives
$$
|f(h)-f(0)-L(h)|\le\frac{\varepsilon}{\sqrt p}\sum_j|h_j|\le\varepsilon\|h\|_2.
$$
Thus $f$ is <Fréchet differentiable> at zero, with <derivative> $L$. No continuity of the <partial derivatives> away from zero was used.
For the supplied one-dimensional function,
$$
g'(s)=\begin{cases}2s\sin(1/s)-\cos(1/s),&s\ne0,\\0,&s=0.\end{cases}
$$
The derivative at zero follows from $g(s)/s=s\sin(1/s)\to0$. For nonzero $s$ the derivative is continuous. At zero, the sequences $s_k=1/(2\pi k)$ and $t_k=1/((2k+1)\pi)$ give derivative values $-1$ and $1$. Since $f_x(x,y)=g'(x)$, \b[the partial derivative is continuous exactly where $x\ne0$], irrespective of $y$.
Nevertheless \b[$f$ is differentiable at every point of the plane]. The one-dimensional function $g$ is differentiable at every real number. At a fixed $(x_0,y_0)$,
$$
g(x_0+h)-g(x_0)=g'(x_0)h+o(|h|),\qquad
g(y_0+k)-g(y_0)=g'(y_0)k+o(|k|).
$$
Their sum has remainder $o(\sqrt{h^2+k^2})$, which proves <Fréchet differentiability> with derivative $(h,k)\mapsto g'(x_0)h+g'(y_0)k$. In particular, at the origin $|f(h,k)|\le h^2+k^2$, an immediate remainder bound. Continuity of all partial derivatives is sufficient, not necessary, for differentiability.
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