First problem: Just a typo. Should be dx/dt = 3y^2/(6y). You had "x^2."

Third problem:

You have y' = 4/5 + (3/5)y'

So (5/5)y' - (3/5)y' = 4/5

(2/5)y' = 4/5

You've got an extra negative sign.

Fourth problem:

x^2 + y^3 = 5y

Take the first derivative:

2x + 3y^2*y' = 5 y'

Take the second derivative:

2 + 6y*y' + 3y^2*y'' = 5y''

3y^2*y'' - 5y'' = -2 - 6y*y'

y'' = -(6y*y' + 2)/(3y^2 - 5)

You've got the point (x,y) from the problem and y' at this point from the last problem.

And for the record, you can TRY to bug me all night, but I'm going to be in bed by 11.

-Dan