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A function is Fréchet differentiable at $x$ when there is a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThis 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 giveswhich 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, $q\ne 0$ and the formula is a composition of smooth functions, hence differentiable. At the origin, $|g_1(x,y)|\le |x^2-y^2|\le x^2+y^2$, 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 $a\ne 0$. Along the $x$ direction,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. ConsequentlyFor $g_2$, the formula is smooth away from zero. At zero, $|g_2(x,y)|\le x^2+y^2$, so the same remainder estimate gives zero derivative. HenceNo limit of the sine factor itself is needed for either derivative at the origin.