$解：设y_1=k_1(x+3)，y_2=\frac{k_2}{x^2}(k_1≠0，k_2≠0).$
$∵y=y_1-y_2，$
$∴y=k_1(x+3)-\frac{k_2}{x^2}.$
$根据题意，得\left\{ \begin{array}{l}{4k_1-k_2=-2}\\ {-\frac{k_2}{9}=2}\ \end{array} \right.$
$解得\left\{ \begin{array}{l}{k_1=-5}\\ {k_2=-18}\ \end{array} \right.$
$∴y=-5(x+3)+\frac{18}{x^2}$
$∴当x=-1时，y=-5×2+18=8$
