yep, yall are handy... love ya's much.
So, it asks for us to find the y'' of x2 + xy + y3 = 1 implicitly.
Pointers?
Yes. Rather than do that problem for you, let me find the second derivative of y when $\displaystyle y^2- e^x+ xy- 2y= 4$.
Differentiating once, $\displaystyle 2yy'- e^x+ y+ xy'- 2y'= 0$. Differentiating again, $\displaystyle 2y'y'+ 2yy''- e^x+ y'+ y'+ xy''- 2y''= 2(y')^2+ 2yy''- e^x+ 2y'+ xy''- 2y''= 0$. Now, you can solve for y'':
$\displaystyle y''= \frac{e^x- 2(y')^2- 2y'}{2y+ x- 2}$.
You could, now, replace y' from the first derivative but typically, that is not necessary.
as i retry this... Im workin on finding the formula for dy/dx
and i got to here... 3x^2+[x(dy/dx) + y(dx/dx)]+2y=0
now, the stuff in the brackets.. how do I simplify that... in the first term can I 'cancel' the x leaving y in the numerator. but then that leaves the second term...
No, you can't cancel inside the brackets. The symbol $\displaystyle dy/dx$ is one symbol. It is not a fraction. However, the $\displaystyle dx/dx = 1$, so that simplifies a bit. Incidentally, I would agree with your differentiation except for the last term. You should use the chain rule on it thus:
$\displaystyle 3x^{2}+[xy'+y]+2yy'=0.$
Can you continue?