For which $y$ is $3 x^4-4 x^2 y+y^2<0$ How does one show that the function $f(x,y)=3 x^4-4 x^2 y+y^2$ is negative for $x^2\lt y&lt;3x^2$?

I do not understand what exactly to do to determine this. I have studies linear inequalities before.

My poor attempt:

$3 x^4-4 x^2 y+y^2&lt;0 \Leftrightarrow y^2-4x^2y&lt;-3x^4 \Leftrightarrow y(y-4x^2)&lt;-3x^4$ 

But i'm not sure how to proceed from here. Any help is appreciated
Hint: $3 x^4-4 x^2 y+y^2=(y-x^2) (y-3 x^2)$.
It's $$(y-2x^2)^2-x^4&lt;0$$ or $$(y-3x^2)(y-x^2)&lt;0$$ or
$$x^2&lt;y&lt;3x^2,$$ which is possible for any $x\neq0$.