= Solution
For fixed $(x_1,x_2)$, apply the given one-variable identity to the <smooth function>
$$
u(t)=f(x_1,tx_2).
$$
The <chain rule> gives
$$
u'(0)=x_2D_2f(x_1,0),
\qquad
u''(t)=x_2^2D_2^2f(x_1,tx_2).
$$
Therefore
$$
f(x_1,x_2)
=f(x_1,0)+x_2D_2f(x_1,0)+x_2^2h(x_1,x_2),
$$
where
$$
\boxed{
h(x_1,x_2)
=\int_0^1(1-t)D_2^2f(x_1,tx_2)\,dt
}.
$$
Every derivative of the integrand is continuous. Repeated <differentiation under the integral sign> therefore shows that $h$ is smooth. This is the <Second-order Hadamard lemma>.
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