 Solution

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.