= Solution
The function is <Fréchet differentiable> at $(a,b)$ when there is a <linear map> $L:\mathbb R^2\to\mathbb R$ such that
$$
f(a+h,b+k)=f(a,b)+L(h,k)+r(h,k),\qquad
\lim_{(h,k)\to(0,0)}\frac{|r(h,k)|}{\sqrt{h^2+k^2}}=0.
$$
If it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.
To prove <differentiability from continuous partial derivatives>, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable <mean value theorem> gives
$$
\begin{aligned}
f(a+h,b+k)-f(a,b)
&=f(a+h,b+k)-f(a,b+k)+f(a,b+k)-f(a,b)\\
&=h f_x(a+\theta h,b+k)+k f_y(a,b+\eta k),
\end{aligned}
$$
for some $0<\theta,\eta<1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding <partial derivative> on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both <partial derivatives> at $(a,b)$, is at most
$$
\epsilon(|h|+|k|)\le\sqrt2\epsilon\sqrt{h^2+k^2}
$$
for sufficiently small increments. Since $\epsilon$ is arbitrary, the normalized remainder tends to zero. Thus \b[the function is differentiable at $(a,b)$], with the displayed derivative.
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