Differentiating $\frac{dy}{dx}=3x+2y+xy$? What would $y''$ be of $\frac{dy}{dx}=3x+2y+xy$? In other words, what is the result ($y''$) if I differentiate $y'=3x+2y+xy$? 
Can I do the following:
Set $$f(x,y) = 3x+2y+xy$$
then
$$f'(x,y) = \frac{df}{dx} + \frac{df}{dy}\frac{dy}{dx}$$
$$ f'(x,y) = (3 + y)+(2+x)(y')$$
$$ f'(x,y) = 3 + y + 2y'+ xy'$$
?

$\frac {dy}{dx} = 3x + 2y + xy$
Differentiate both sides using the rules of implicit differentiation:
$\frac{d}{dx}\frac {dy}{dx} = \frac{d}{dx} (3x + 2y + xy)\\
\frac {d^2y}{dx^2} = \frac{d}{dx} 3x + \frac{d}{dx}2y + \frac{d}{dx}xy\\
\frac {d^2y}{dx^2} = 3 + 2\frac {dy}{dx} + y + x\frac{dy}{dx}\\
$
We can substitute $\frac {dy}{dx} = 3x + 2y + xy$ from the begining
$\frac {d^2y}{dx^2} = 3 + y + (2+x)(3x + 2y + xy)$
And simplify.
