Let $T:X \to Y$ be a nonlinear operator between Hilbert spaces. Is it possible for $h \mapsto T'(x)[h]$ to be a compact mapping from $X$ to $Y$ even if $T$ is not compact? Here $T'(x)[h]$ is the directional derivative of $T$ at $x$ in the direction $h$.
If so, is it possible to classify all such $T$?
I do not know the answer to how to classify all such $T$, but one example is the following.
Let $X=Y=\ell^2$ and $T$ be the squaring operator $$ (Tx)_n=x_n^2. $$ Since $Te_k=e_k$, the operator is clearly not compact. However, it is Fréchet differentiable with the derivative being the multiplication operator by $2x$ $$ T'(x)[h]=2\begin{bmatrix}x_1 &&&\\&x_2&&\\&&x_3&\\&&&\ddots\end{bmatrix}\begin{bmatrix}h_1\\h_2\\h_3\\\vdots\end{bmatrix}. $$ This operator is compact, because the finite truncations converge to it in the operator norm.