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.

Alternative method:
Write $f(x)$ as $\frac{x-1}{(x-1)^2+2}$. Thus, your function is a horizontal shift of $g(t)=\frac{t}{t^2+2}$, and hence these functions have the same range.
Next, $g$ is odd, so it is sufficient to study positive $t$ (for which $g$ is positive). Now, by the arithmetic-geometric mean $t^2+2\geq 2\sqrt{t^2}\sqrt{2}=2\sqrt{2}t$ with equality when $t^2=2$. Thus the maximum value is $\frac{1}{2\sqrt{2}}$. With symmetry, the minimum is $-\frac{1}{2\sqrt{2}}$. Thus, the range is (by continuity) $[-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}]$.
