 Solution

Differentiability at $p$ means $f(p+h)=f(p)+Lh+o(|h|)$ for a linear map $L=Df(p)$; partial derivatives are its values on coordinate vectors. Integrating the partial derivatives along the two coordinate segments and using continuity at $p$ proves the stated sufficient condition. Here $f(x,y)=g(x)+g(y)$ with $g(t)=t^2\sin (1/t)$ and $g(0)=0$. Since $g\prime (0)=0$, $f$ is differentiable everywhere. But $g\prime (t)=2t\sin (1/t)-\cos (1/t)$ does not approach $0$ at zero, so $f$ is $C^1$ exactly where $xy\ne 0$.