= Solution
<Differentiability> at $(a,b)$ means that there is a <linear map> $L:\mathbb R^2\to\mathbb R$ such that
$$
f(a+h,b+k)-f(a,b)=L(h,k)+o(\sqrt{h^2+k^2}).
$$
Necessarily $L(h,k)=f_x(a,b)h+f_y(a,b)k$ when those <partial derivatives> exist. Under the stated assumptions, apply the one-variable <mean value theorem> separately to the two coordinate increments:
$$
f(a+h,b+k)-f(a,b)=h f_x(a+\theta h,b+k)+k f_y(a,b+\eta k),
$$
for some intermediate $\theta,\eta\in(0,1)$, omitting a term if its increment is zero. Existence of the <partial derivatives> in a neighborhood supplies the differentiability along each segment required for that theorem. Their <continuity> at $(a,b)$ makes the difference from $f_x(a,b)h+f_y(a,b)k$ bounded by $(|h|+|k|)\varepsilon(h,k)$, where $\varepsilon(h,k)\to0$. Since $|h|+|k|\le\sqrt2\sqrt{h^2+k^2}$, this is the required little-o remainder and proves <Fréchet differentiability>.
For the particular function, $f(h,0)=f(0,k)=0$, so the defining difference quotients give \b[$f_x(0,0)=f_y(0,0)=0$]. However $f(t,t)=1/2$ for every $t\ne0$. It is not continuous at the origin, and therefore \b[not differentiable there]. This illustrates why existence of <partial derivatives> alone is insufficient.
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