Find $y''=y''(x)$ for the function defined by equation $xy+y^2=2x$
I got the first derivative just fine, $y'(x)= \frac{2-y(x)}{x+2y(x)}$ but I'm having trouble with finding the second derivative. After differentiating again and inserting the first derivative the expression becomes quite complicated and hard to simplify. Is there a way to check my answer? This is what I got after trying to simplify it: $y''(x)=\frac{2y(x)^2+2xy(x)-4x-8}{(x+2y(x))^3}$
by differentiating we obtain $y+xy'+2yy'=2$ thus we get $y'=\frac{2-y}{x+2y}$ and so we get $y''=\frac{-y'(x+2y)-(2-y)(1+2y')}{(x+2y)^2}$
$y''=\frac{-xy'-2+y-4y'}{(x+2y)^2}$