jmc

Since there are vertical asymptotes at $x=-3$ and $x=2,$ we can assume that $q(x)=(x+3)(x-2).$

Since the graph passes through $(0,0),$ $p(x)=kx$ for some constant $k.$ Thus, $$\frac{p(x)}{q(x)}=\frac{kx}{(x+3)(x-2)}.$$To find $k,$ note that the graph passes through $(3,1).$ Thus, $$\frac{3k}{(6)(1)}=1.$$Hence, $k=2,$ and $$\frac{p(x)}{q(x)}=\frac{2x}{(x+3)(x-2)}.$$Then $$\frac{p(-1)}{q(-1)}=\frac{2(-1)}{(2)(-3)}=\boxed{\frac{1}{3}}.$$