$\displaystyle y^2=cos(xy)+ln(x+y)$

Find $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$

Now, this is what I did, I'm not entirely sure if I'm doing it right.

$\displaystyle 2y.\frac{dy}{dx} = {\color{blue}-ysin(xy).(y+x\frac{dy}{dx})}+{\color{red}\frac{1}{ x} + \frac{1}{y}\frac{dy}{dx}}$

$\displaystyle \frac{-1}{x}+ysin(xy)=(\frac{1}{y}-xsin(xy)-2y)\frac{dy}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{ysin(xy)-\frac{1}{x}}{\frac{1}{xy}-xsin(xy)-2y}$

So do i get

or

?