Let f(x,y) = x^5+3x^4y-5x^3y^2-15x^2y^3+4xy^4+12y^5

Substitute x=y to f(x), then f(x) = 0

which means (x-y) is one of factors of f(x)

Group the terms of f(x) to factor out x-y

x^5+3x^4y-5x^3y^2-15x^2y^3+4xy^4+12y^5

=(x^5-x^4y) + (4x^4y - 4x^3y^2) -(x^3y^2-x^2y^3)-(16x^2y^3-16xy^4)-(12xy^4-12y^5)

=(x-y)( x^4+4x^3y-x^2y^2-16xy^3-12y^4)

Let g(x) = x^4+4x^3y-x^2y^2-16xy^3-12y^4

Substitute x=2y to g(x), then g(x) = 0

which means (x-2y) is one of factors of g(x)

Group the terms of g(x) to factor out x-2y

x^4+4x^3y-x^2y^2-16xy^3-12y^4

(x^4-16y^4)+(4x^3y -16xy^3)-(x^2y^2-4y^4)

=(x^2+4y^2)(x^2-4y^2)+4xy(x^2-4y^2)-y^2(x^2-4y^2)

=(x^2+4y^2+4xy-y^2)(x^2-4y^2)

=( x^2+4xy +3y^2)(x+2y)(x-2y)

=(x+y)(x+3y)(x+2y)(x-2y)

In summary, the expression could be factorized to the product of 5 terms

x^5+3x^4y-5x^3y^2-15x^2y^3+4xy^4+12y^5

=(x-y)(x+y)(x+3y)(x+2y)(x-2y)

If x =\pm y, x=-3y or x=\pm2y, the expression is equal to 0, not 33

Otherwise, the expression could be factorized at least 4 different integers other than 1 and itself. But the factors of 33 are just 3 and 11 other than 1 and itself.

Therefore,

It's not possible for x^5+3x^4y-5x^3y^2-15x^2y^3+4xy^4+12y^5 to be equal to 33