= Solution
The <Fréchet derivative> at $p=(x_0,y_0)$ is a <linear map> $L:\mathbb R^2\to\mathbb R$ for which
$$
f(p+h)=f(p)+Lh+o(\|h\|)\quad(h\to0).
$$
This is <differentiability>; if it holds, $L(h_1,h_2)=D_1f(p)h_1+D_2f(p)h_2$.
To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:
$$
f(x_0+h,y_0+k)-f(p)
=[f(x_0+h,y_0+k)-f(x_0,y_0+k)]
+[f(x_0,y_0+k)-f(p)].
$$
The one-variable <mean value theorem>, applicable because the restrictions are differentiable, represents these two terms as
$$
hD_1f(x_0+\theta h,y_0+k)+kD_2f(x_0,y_0+\phi k),\qquad0<\theta,\phi<1.
$$
The zero-increment cases are interpreted directly. Subtract $D_1f(p)h+D_2f(p)k$. By <continuity> of both <partial derivatives> at $p$, the remainder is at most $\varepsilon(|h|+|k|)\leq\sqrt2\varepsilon\sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves <differentiability>.
The converse is false. Set $f(x,y)=x^2\sin(1/x)$ for $x\ne0$, and $f(0,y)=0$. At each point with $x=0$, its <Fréchet derivative> is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. But
$$
D_1f(x,y)=2x\sin(1/x)-\cos(1/x)\quad(x\ne0),\qquad D_1f(0,y)=0,
$$
is not continuous there. Thus a function can be differentiable everywhere while its <partial derivatives> are discontinuous.
For $h$, every point with $x\ne0$ is a smooth point. At the origin, $|h(x,y)|\leq|xy|\leq(x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne0$, the putative first <partial derivative> would require the limit $y_0\sin(1/x)$, which does not exist. Consequently the exact set of differentiability is
$$
\boxed{\{(x,y):x\ne0\}\ \cup\ \{(0,0)\}.}
$$
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