Why do you believe it can be solved?
What methods have you for your attempt?
What have you tried?
Just looking at it, I think I'd try y = 1 and y = -1.
Let x = y^(1/3), so y^(2/3) = x^2 and y^4 = y^(12/3) = x^12.
Substituting, you should get x = 2 - x^12, so you are trying to find the solutions of x^12 + x^2 - 2 = 0, which are the x-intercepts of f = x^12 + x^2 - 2.
Plug x^12 + x^2 - 2 into your graphing calculator to "see" where the two real-values zeroes are. Divide out the two rational zeros (which duplicate the two rational zeroes of the original equation).
You should be left with an equation in x^10 which has no real zeroes.