For which $y$ is $3 x^4-4 x^2 y+y^2<0$

Question

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

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

My poor attempt:

$3 x^4-4 x^2 y+y^2<0 \Leftrightarrow y^2-4x^2y<-3x^4 \Leftrightarrow y(y-4x^2)<-3x^4$

But i'm not sure how to proceed from here. Any help is appreciated

Answer 1

It's $$(y-2x^2)^2-x^4<0$$ or $$(y-3x^2)(y-x^2)<0$$ or $$x^2<y<3x^2,$$ which is possible for any $x\neq0$.

Answer 2

Hint: $3 x^4-4 x^2 y+y^2=(y-x^2) (y-3 x^2)$.

Answer 3

Hint. That's quadratic in $x^2$ and in $y.$ Make the substitution $t=x^2$ and see if you can factorize.