Given that 

$3 \sin xy + 4\cos xy = 5$

Let $xy = t$

$x \frac{dy}{dx} + y = \frac{dt}{dx}----\left(I\right)$

$\Rightarrow 3 \sin t +4 \cos t = 5$

Differentiate on both sides

$\Rightarrow \left(3 \cos t - 4 \sin t\right)\frac{dt}{dx}=0$

$3 \cos t -4 \sin t \ne 0 \Rightarrow \frac{dt}{dx} = 0$

$x\frac{dy}{dx} + y =0$

$\Rightarrow \frac{dy}{dx} = \raisebox{1ex}{$-y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.$.