To find the values of a, b, and c such that the maximum value of the directional derivative is in the direction parallel to the y-axis and has a magnitude of 6, we can follow these steps:

1. Determine the gradient of the function:
The gradient of f(x, y, z) = axy^2 + byz + cx^2z^2 is given by:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

Taking the partial derivatives of f(x, y, z) with respect to each variable, we get:
∂f/∂x = 2acxz^2
∂f/∂y = 2axy
∂f/∂z = by + 2cx^2z

2. Find the direction parallel to the y-axis:
Since we want the maximum value of the directional derivative in the direction parallel to the y-axis, the direction vector will be (0, 1, 0).

3. Calculate the directional derivative:
The directional derivative of f(x, y, z) in the direction of the vector v = (0, 1, 0) is given by:
Dv(f) = ∇f · v

Substituting the partial derivatives and the direction vector, we get:
Dv(f) = (∂f/∂x, ∂f/∂y, ∂f/∂z) · (0, 1, 0)
= 2axy

4. Set up the equation for the magnitude of the directional derivative:
We want the magnitude of the directional derivative to be 6, so we have:
|Dv(f)| = |2axy| = 6

5. Solve for the values of a and b:
From the equation |2axy| = 6, we can divide both sides by 2xy:
|a| = 3/|xy|

Since we want the maximum value, we can assume a = 3/|xy|. Substituting this value into the equation for the directional derivative, we get:
Dv(f) = 2(3/|xy|)xy
= 6

Thus, a = 3/|xy|.

6. Solve for the value of c:
From the partial derivative ∂f/∂z = by + 2cx^2z, we can substitute the values of a = 3/|xy| and Dv(f) = 6:
by + 2cx^2z = 6

Since we want the maximum value, we can assume b = 0. Substituting this value into the equation, we get:
2cx^2z = 6

Thus, c = 3/(x^2z).

In summary:
- The value of a is 3/|xy|.
- The value of b is 0.
- The value of c is 3/(x^2z).

Please note that these values may vary depending on the specific values of x and z, as they were not given in the question.