First derivative:

xy' + y + 2yy' = 0

[I am using y' instead of dy/dx. It is easier to type and quite easy to read.]

I have used the product rule (for the product xy) and the chain rule (for y²).

Second derivative:

[xy'' + y'] + [ y'] + [2yy'' + y'(2y')] = 0

and this simplifies to:

(x + 2y)y'' + 2y' + 2(y')² = 0

We can solve for y'':

y'' = [-2y' − 2(y')²] / (x + 2y)

Here's a movie giving a different view of this example. It uses:

1. dy/dx notation;
2. A different approach to the problem (in the movie, we find the expression for dy/dx first, then differentiate that to get the second derivative). The result is a simpler expression for the second derivative.

The answer looks quite different, but it does have the same value.