It's pouring down rain, and the amount of rain hitting point $(x,y)$ is given by 
$$f(x,y)=|x^3+2x^2y-5xy^2-6y^3|.$$
If you start at the origin $(0,0)$, find all the possibilities for $m$ such that $y=mx$ is a straight line along which you could walk without any rain falling on you.
1. We start with the given function:
   \[
   f(x,y) = |x^3 + 2x^2y - 5xy^2 - 6y^3|
   \]
   We need to find the values of \( m \) such that \( y = mx \) results in \( f(x, mx) = 0 \).

2. Substitute \( y = mx \) into the function:
   \[
   f(x, mx) = |x^3 + 2x^2(mx) - 5x(mx)^2 - 6(mx)^3|
   \]
   Simplify the expression inside the absolute value:
   \[
   f(x, mx) = |x^3 + 2mx^3 - 5m^2x^3 - 6m^3x^3|
   \]
   Factor out \( x^3 \):
   \[
   f(x, mx) = |x^3 (1 + 2m - 5m^2 - 6m^3)|
   \]
   For \( f(x, mx) = 0 \), the expression inside the absolute value must be zero:
   \[
   1 + 2m - 5m^2 - 6m^3 = 0
   \]

3. Solve the cubic equation \( 1 + 2m - 5m^2 - 6m^3 = 0 \):
   - Notice that \( m = -1 \) is a root because substituting \( m = -1 \) gives:
     \[
     1 + 2(-1) - 5(-1)^2 - 6(-1)^3 = 1 - 2 - 5 + 6 = 0
     \]
   - Factor \( m + 1 \) out of the cubic polynomial:
     \[
     1 + 2m - 5m^2 - 6m^3 = (m + 1)(-6m^2 - 5m + 1)
     \]
   - Solve the quadratic equation \( -6m^2 - 5m + 1 = 0 \) using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     m = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-6)(1)}}{2(-6)} = \frac{5 \pm \sqrt{25 + 24}}{-12} = \frac{5 \pm \sqrt{49}}{-12} = \frac{5 \pm 7}{-12}
     \]
     This gives two solutions:
     \[
     m = \frac{5 + 7}{-12} = \frac{12}{-12} = -1
     \]
     \[
     m = \frac{5 - 7}{-12} = \frac{-2}{-12} = \frac{1}{6}
     \]

4. The complete set of solutions for \( m \) is:
   \[
   m = -1, \frac{1}{2}, -\frac{1}{3}
   \]

The final answer is \(\boxed{-1, \frac{1}{2}, -\frac{1}{3}}\)