$解:(1)设y_1=k_1x,y_2=\frac{k_2}{x-2},∵y=y_1+y_2,∴y=k_1x+\frac{k_2}{x-2}$
$将x=1,y=2和x=3,y=10分别代入y=k_1x+\frac {k_2}{x-2}中,$
$得\begin{cases}{2=k_1-k_2,}\\{10=3k_1+k_2,}\end{cases}解得\begin{cases}{k_1=3,}\\{k_2=1,}\end{cases}\ $
$∴y与x的函数表达式为y=3x+\frac {1}{x-2}$
$(2)将x=-1代入y=3x+\frac {1}{x-2},得y=-\frac{10}{3},∴当x=-1时,y的值为-\frac{10}{3}.$