Step 1: Decide what controls increasing or decreasing.
For $y = \frac{u}{v}$, the sign of $y'$ tells the story. If $y' < 0$ everywhere, the function decreases everywhere.
Step 2: Name the parts.
Let $u = \sin x + 2\cos x$ and $v = 3\sin x + 4\cos x$. Then $u' = \cos x - 2\sin x$ and $v' = 3\cos x - 4\sin x$.
Step 3: Use the quotient rule and focus on the numerator.
$y' = \frac{u'v - uv'}{v^2}$. The denominator $v^2$ is always positive, so the sign of $y'$ is the sign of $N = u'v - uv'$.
Step 4: Expand the numerator.
Multiply out $u'v = (\cos x - 2\sin x)(3\sin x + 4\cos x)$ and $uv' = (\sin x + 2\cos x)(3\cos x - 4\sin x)$. After expanding and subtracting, the cross terms with $\sin x\cos x$ cancel.
Step 5: Simplify using $\sin^2 x + \cos^2 x = 1$.
The numerator collapses to a constant. Working it through gives $N = -2(\sin^2 x + \cos^2 x) = -2$.
Step 6: Read the conclusion.
Since $N = -2 < 0$ for every $x$, $y' < 0$ for all real $x$. So the function decreases for all $x \in \mathbb{R}$. \[ \boxed{\text{decreases for all } x \in \mathbb{R}} \]