= Solution
A function is <Fréchet differentiable> at $x$ when there is a <linear map> $A:\mathbb R^n\to\mathbb R^m$ such that
$$
f(x+h)=f(x)+Ah+r(h),\qquad\frac{|r(h)|}{|h|}\longrightarrow0.
$$
This <linear map>, which is unique, is the <derivative> $Df(x)$.
For the <chain rule>, if $f$ is <differentiable> at $x$ and $g$ is <differentiable> at $f(x)$, then $g\circ f$ is <differentiable> at $x$ and $D(g\circ f)(x)=Dg(f(x))Df(x)$. Indeed write $f(x+h)-f(x)=Ah+r(h)=v(h)$ and $g(f(x)+v)=g(f(x))+Bv+s(v)$, where $r(h)=o(|h|)$, $s(v)=o(|v|)$. Since $v(h)=O(|h|)$, substitution gives
$$
g(f(x+h))-g(f(x))=BAh+Br(h)+s(v(h))=BAh+o(|h|),
$$
which proves the assertion, including paths on which $v(h)=0$.
For $g_1$, write $q=x^2-y^2$. Away from the two diagonal lines, $q\ne0$ and the formula is a composition of smooth functions, hence <differentiable>. At the origin, $|g_1(x,y)|\leq|x^2-y^2|\leq x^2+y^2$, so its ratio to the displacement length tends to zero. Its <derivative> there is therefore zero: this is <quadratic damping restores differentiability at the origin>.
At a nonzero point $(a,b)$ with $a=\pm b$, necessarily $a\ne0$. Along the $x$ direction,
$$
\frac{g_1(a+t,b)-g_1(a,b)}t
=(2a+t)\sin\frac1{2at+t^2}.
$$
The denominator inside the sine is locally an invertible function of $t$, with nonzero <derivative> $2a$. Choose sequences tending to zero for which its reciprocal equals $2\pi n+\pi/2$ and $2\pi n+3\pi/2$. The quotients tend to $2a$ and $-2a$. Thus even this <directional derivative> fails to exist, and the function is not <differentiable> there. Consequently
$$
\boxed{\operatorname{Diff}(g_1)=\{(x,y):x\ne y,\ x\ne-y\}\cup\{(0,0)\}.}
$$
For $g_2$, the formula is smooth away from zero. At zero, $|g_2(x,y)|\leq x^2+y^2$, so the same remainder estimate gives zero <derivative>. Hence
$$
\boxed{\operatorname{Diff}(g_2)=\mathbb R^2,\qquad Dg_2(0,0)=0.}
$$
No limit of the sine factor itself is needed for either <derivative> at the origin.
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