Answer
All the real numbers except $x=-2$ and $x=-4$.
Work Step by Step
$\frac{x^2-5x+6}{x^2+6x+8}$
The denominator can not be equal to zero:
$x^2+6x+8\ne0$
$x^2+4x+2x+8\ne0$
$x(x+4)+2(x+4)\ne0$
$(x+2)(x+4)\ne0$
$x\ne-2$ and $x\ne-4$