= Solution
A function $f:U\to\mathbb R^n$ has a <Fréchet derivative> at $a\in U$ if there is a <linear map> $A:\mathbb R^m\to\mathbb R^n$ such that
$$
\boxed{f(a+h)=f(a)+Ah+o(\|h\|)\quad(h\to0)}.
$$
The 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 yields
$$
f(a_1+h,a_2+k)-f(a)
=h f_x(a_1+\theta h,a_2+k)+k f_y(a_1,a_2+\sigma k)
$$
for $\theta,\sigma\in(0,1)$, 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 $\varepsilon>0$ the remainder has modulus at most $\varepsilon(|h|+|k|)\leq\sqrt2\varepsilon\sqrt{h^2+k^2}$ 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 <Fréchet derivative> there would therefore be $A(h,k)=h$. However,
$$
f(t,t)=\frac t2+t^2,\qquad
\frac{|f(t,t)-t|}{\sqrt2|t|}\longrightarrow\frac1{2\sqrt2}\ne0.
$$
Thus
$$
\boxed{f\text{ is differentiable exactly on }\mathbb R^2\setminus\{(0,0)\}}.
$$
The failure is not merely a failure of <continuity>: $|f(x,y)|\leq |x|+2y^2\to0$ at the origin, so the function is <continuous> there.
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