For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:

Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$


I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{3})$.
But the teacher said I should use another way.
I didn't find any.

The universe is the set of real numbers.
$\forall x\exists y<1:3x^2y+y-x=0$
$\iff\forall x\exists y<1:3x^2y+y=x$
$\iff\forall x\exists y<1:y(3x^2+1)=x$
$\iff\forall x\exists y<1:y=\frac{x}{3x^2+1}$, because we know that $3x^2+1\not=0$
$\iff\forall x\exists y(y<1\land y=\frac{x}{3x^2+1})$
$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land y<1)$
$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land \frac{x}{3x^2+1}<1)$
$\iff\forall x(\exists y(y=\frac{x}{3x^2+1})\land \frac{x}{3x^2+1}<1)$
$\iff\forall x\frac{x}{3x^2+1}<1$
$\iff\forall x:x<3x^2+1$, because we know that $3x^2+1>0$
$\iff\forall x:3x^2-x+1>0$
$\iff\forall x:x^2-\frac{1}{3}x+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2-(\frac{1}{6})^2+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2-\frac{1}{36}+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2>\frac{-11}{36}$, which is true, because the square of every real number is positive.
