= Solution
Put $W=\sqrt{1+h_x^2+h_y^2}$. An upward-pointing Gauss map is
$$
N=\frac{(-h_x,-h_y,1)}{W}.
$$
For <tangent vectors> $v,w$, <differentiation> of
$$
d\phi_t=d\phi+t\,dN
$$
gives
$$
\left.\frac d{dt}\right|_{t=0}I^{S_t}(v,w)
=\langle dN(v),d\phi(w)\rangle
+\langle d\phi(v),dN(w)\rangle.
$$
The Weingarten map is self-adjoint and $S=-dN$, so the right side is $-2II^S(v,w)$. Hence
$$
II^S=-\frac12\left.\frac d{dt}\right|_{t=0}I^{S_t}.
$$
Alternatively differentiating $\langle N,\phi_i\rangle=0$ gives
$$
II_{ij}=-\langle N_i,\phi_j\rangle
=\langle N,\phi_{ij}\rangle.
$$
Therefore
$$
\boxed{
II^S=\frac1W
\begin{pmatrix}
h_{xx}&h_{xy}\\
h_{xy}&h_{yy}
\end{pmatrix}.}
$$
This is the <fundamental forms of a graph surface>.
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