Find $y'$ for $\ln(x+y)=\arctan(xy)$ 
Find $y'$ for $\ln(x+y)=\arctan(xy)$

Here is my attempt at a solution. Is this correct? Any hints or advice would be appreciated.


Even if the result is correct, I guess that a systematic use of total differentiation makes things slightly easier.
Consider $$F=\log(x+y)-\tan^{-1}(xy)=0$$ So $F'_x~dx+F'_y~dy=0$. I this case, $$F'_x=\frac{1}{x+y}-\frac{y}{x^2 y^2+1}$$ $$F'_y=\frac{1}{x+y}-\frac{x}{x^2 y^2+1}$$ $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{\frac{y}{x^2 y^2+1}-\frac{1}{x+y}}{\frac{1}{x+y}-\frac{x}{x^2 y^2+1}}=\frac{y \left(x+y-x^2y\right)-1}{x^2 \left(y^2-1\right)-x y+1}=- \frac{(x+y)y-(1+x^2y^2)}{(x+y)x-(1+x^2y^2)}$$
