Range of $f(x) =\frac {x -1}{x^2 -2x + 3} $? Is my solution for finding the range of 
$$f(x) = \frac{x-1}{x^2 -2x + 3} $$
correct?
Since its Domain is $ \mathbb{R} $, so transforming this equation into $x$ in terms of $y$ , we get $$ yx^2 - (2y +1)x + (3y+1) = 0 $$
Now, since $ x \in \mathbb {R} $ , so its discriminant $\mathbb{D}$ must be $ \ge 0 $, so we have, $$ (2y + 1)^2 - 4y(3y + 1) \ge 0 $$
Solving it, we get
$$ y \in \left[-{1 \over {2 \sqrt2}}, {1 \over {2 \sqrt2}}\right] $$
Kindly solve my problem.

Everything looks good, except you have a small mistake in your equation. Edit: There is no mistake anymore after OP edited his question
It should be $y x^2 - (2y + 1)x + 3y - 1 = 0$. This is solvable if and only if $$(2y + 1)^2 - 4y(3y - 1) = -8y^2 + 8y + 1 \ge 0.$$
The roots of $-8y^2 + 8y + 1 = 0$ are at
$$\frac{2 \pm \sqrt{6}}{4},$$
so we need to have $y \in \left[\frac{2 - \sqrt{6}}{4}, \frac{2 + \sqrt{6}}{4}\right]$.
