You should consider the following substitution, such that:

`x = 0 => f(0 + y) - f(0 - y) = 2f(y) + 6*0*y`

`f(y) - f(-y) = 2f(y)`

You need to combine duplicate terms such that:

`-f(-y) = 2f(y) - f(y) => f(-y) = -f(y)`

Since the statement `f(-y) = -f(y)` represents an odd function, hence, if x = 0, the function is odd.

You should consider the following replacements x = y and y = x, such that:

`f(y + x) - f(y - x) = 2f(x) + 6y^2x => f(x + y) + f(x - y) = 2f(x) + 6xy^2`

You need to add this relation to the relation provided by the problem, such that:

`f(x + y) - f(x - y) + f(x + y) + f(x - y)= 2f(y) + 2f(x) + 6x^2y + 6xy^2`

Reducing duplicate terms yields:

`2f(x + y) = 2(f(x) + f(y) + 3x^2y + 3xy^2)`

`f(x + y) = (f(x) + f(y) + 3xy(x + y))`

You should use the following special product, such that:

`(x + y)^3 = x^3 + y^3 + 3xy(x + y) => 3xy(x + y) = (x + y)^3 - x^3 - y^3`

You may replace `(x + y)^3 - x^3 - y^3` for `3xy(x + y)` such that:

`f(x + y) = f(x) + f(y) + (x + y)^3 - x^3 - y^3`

`f(x + y) - (x + y)^3 = f(x) - x^3 + f(y) - y^3`

You may use the following function, such that:

`g(x) = f(x) - x^3`

Reasoning by analogy yields:

`g(y) = f(y) - y^3`

`g(x + y) = g(x + y) - (x + y)^3`

Hence, using the functions `g(x), g(y)` and `g(x+y)` , the relation `f(x + y) - (x + y)^3 = f(x) - x^3 + f(y) - y^3` may be converted in the following relation, such that:

`g(x+y) = g(x) + g(y)`

Since `g(x+y) = g(x) + g(y)` represents the Cauchy's functional equation whose solution are `g(x) = alpha*x` yields:

`f(x) = x^3 + alpha x`

**Hence, evaluating the function `f(x)` , under the given conditions, yields `f(x) = x^3 + alpha x` .**