 Solution

The Frechet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ 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 $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Frechet 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. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, 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