Answer
$$ y'(x) =\frac{y-1+3 \sin (3 x-y)}{\sin (3 x-y)-x}$$
Work Step by Step
Given $$x+\cos (3 x-y)=x y$$
Differentiate with respect to $x$
\begin{align*}
1-3 \sin (3 x-y)+y^{\prime} \sin (3 x-y)&=y+x y^{\prime}\\
y^{\prime}[\sin (3 x-y)-x]&=y-1+3 \sin (3 x-y)
\end{align*}
Then
$$ y'(x) =\frac{y-1+3 \sin (3 x-y)}{\sin (3 x-y)-x}$$
