Find the set of values taken by y = 1/(x²+6x+5) for real values of x.

The given equation is $y = \frac{1}{x^2+6x+5}$.
Assuming $y \ne 0$, we can rewrite this as:

$$x^2+6x+5 = \frac{1}{y}$$

Rearranging into the standard quadratic form $ax^2+bx+c=0$:

$$x^2+6x+\left(5-\frac{1}{y}\right) = 0$$

The quadratic equation is $x^2+6x+\left(5-\frac{1}{y}\right) = 0$, where $a=1$, $b=6$, and $c=5-\frac{1}{y}$.
The discriminant $\Delta$ is given by the formula $\Delta = b^2-4ac$.

$$\Delta = (6)^2 - 4(1)\left(5-\frac{1}{y}\right)$$ $$\Delta = 36 - 4\left(5-\frac{1}{y}\right)$$ $$\Delta = 36 - 20 + \frac{4}{y}$$ $$\Delta = 16 + \frac{4}{y}$$

For real values of $x$, we must have $\Delta \ge 0$.

$$16 + \frac{4}{y} \ge 0$$

Combine the terms on the left side by finding a common denominator:

$$\frac{16y}{y} + \frac{4}{y} \ge 0$$ $$\frac{16y+4}{y} \ge 0$$

Factor out 4 from the numerator:

$$\frac{4(4y+1)}{y} \ge 0$$

The critical points are where the numerator is zero ($4y+1=0 \implies y = -\frac{1}{4}$) and where the denominator is zero ($y=0$).
We must exclude $y=0$ because it makes the original function undefined.

We analyze the sign of the expression $\frac{4(4y+1)}{y}$ in the intervals defined by the critical points:

• For $y < -\frac{1}{4}$ (e.g., $y=-1$): $\frac{4(4(-1)+1)}{-1} = \frac{4(-3)}{-1} = 12 \ge 0$. This interval is valid.
• For $-\frac{1}{4} < y < 0$ (e.g., $y=-0.1$): $\frac{4(4(-0.1)+1)}{-0.1} = \frac{4(0.6)}{-0.1} = -24 < 0$. This interval is not valid.
• For $y > 0$ (e.g., $y=1$): $\frac{4(4(1)+1)}{1} = \frac{4(5)}{1} = 20 \ge 0$. This interval is valid.

Since the inequality is $\ge 0$, $y = -\frac{1}{4}$ is included in the solution. $y=0$ is excluded.
Combining the valid intervals, the set of values for $y$ is $(-\infty, -\frac{1}{4}] \cup (0, \infty)$.

The set of values taken by $y$ is $\boxed{\left(-\infty, -\frac{1}{4}\right] \cup (0, \infty)}$.