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
\[\frac{p(-1)}{q(-1)} = \frac{2(-1)}{(2)(-3)} = \boxed{\frac{1}{3}}.\]
The graph of $y = \frac{p(x)}{q(x)}$ is shown below, where $p(x)$ is linear and $q(x)$ is quadratic.  (Assume that the grid lines are at integers.)

[asy]
unitsize(0.6 cm);

real func (real x) {
  return (2*x/((x - 2)*(x + 3)));
}

int i;

for (i = -5; i <= 5; ++i) {
  draw((i,-5)--(i,5),gray(0.7));
  draw((-5,i)--(5,i),gray(0.7));
}

draw((-5,0)--(5,0));
draw((0,-5)--(0,5));
draw((-3,-5)--(-3,5),dashed);
draw((2,-5)--(2,5),dashed);
draw(graph(func,-5,-3.1),red);
draw(graph(func,-2.9,1.9),red);
draw(graph(func,2.1,5),red);

limits((-5,-5),(5,5),Crop);
[/asy]

Find $\frac{p(-1)}{q(-1)}.$
{'answer': '\\frac{1}{3}', 'subject': 'Intermediate Algebra', 'level': 4, 'unique_id': 'train/intermediate_algebra/1671.json'}