Solution
= Solution
{c}
The <Taylor expansions> $\sin t=t+O(t^3)$ and $e^t=1+t+t^2/2+O(t^3)$ give
$$
f(x,y)=-3+6x+3y+\frac12(x+2y)^2+O((|x|+|y|)^3).
$$
Hence the second-order <Taylor polynomial> is
$$
\boxed{T_2(x,y)=-3+6x+3y+\frac12x^2+2xy+2y^2.}
$$
The <sine> contributes no quadratic term at zero.