This is a problem related to differential equations. Let's solve it step-by-step:
1) The provided equation is a first-order nonlinear ordinary differential equation in the form of dy/dx = (x^2 -1) / (3y^2), which can be written as 3y^2 dy/dx = x^2 - 1.
2) Recognizing that this equation is of the method of separable variables because each side of the equation includes only one of the variables (x's and y's). To solve this equation, we will aim to express it in the form Y(y) dy = X(x) dx, which can then be integrated each side separately.
3) Rearrange the equation to have all terms involving x and dx on one side, and all terms involving y and dy on the other: 3y^2 dy = (x^2 - 1) dx
4) The differential equation is now in separable form and we can integrate both sides separately:
The integral of (3y^2) dy = y^3 + C (where C is the constant)
The integral of (x^2 - 1) dx = (x^3/ 3) - x + D (where D is another constant)
5) We can then set both sides of the equation equal to each other to solve for y in terms of x. However, the C and D constants are irrelevant as they will collapse into a single constant E when we subtract one side from the other:
y^3 = (x^3/ 3) - x + E
6) Having got this far, we now need to use the initial condition provided, f(3) = 2, to solve for E:
2^3 = (3^3/ 3) - 3 + E
E = 8 - 9 + 3
E = 2
So,
y^3 = (x^3/ 3) - x + 2
7) Simplifying gives:
y = [(x^3/ 3) - x + 2]^(1/3)
So the particular solution which satisfies the initial condition f(3) = 2 is y = [(x^3/ 3) - x + 2]^(1/3)