Please help. Points of inflection and chain rule

[stanton]

Homework Statement

Skethch the greaph of x^3/(x^3+1). Identify all extrema and points of inflection, asymptote equations, and easily found intercepts

Homework Equations

If a/b=0, a must be 0.(thats how I got critical points from first derivative)
And chain rule: F'(x) = f '(g(x)) g '(x)
And so on.

The Attempt at a Solution

What I have to do is to find points of inflection, asymptotes, and critical points. And I took the first derivative using Quotient rule and got 3x^2/(x^3+1)^2
I am going to take the second derivative of my original function to get ponts of inflection, but I am stuck.

As you can see, f'(x) = 3x^2/(x^3+1)^2
And I pondered a bit for I think I need to use chain rule because of that (x^3+1)^2 in the denominator. But how should I use it? what about the numerator?
I can't just use chain rule in the denominator, and use power rule in nominator on my own, right? I have a bad feeling about this.

This is what I will get if I apply power rule in nominator and apply chain rule in denominator:

6x/[2(x^3+1)(3x^2)] This is wrong, right? because I treated numerator and denominator by using different rules on each of them. Can you check this for me? What should I do in order to get the second derivative?
Or can I just ignore the denominator and just take Quotient rule again to get derivative?

 

[Bohrok]

Your second derivative doesn't look quite right.
Since you're having trouble getting the second derivative, try doing it in steps, starting from f'(x) = 3x²/(x³ + 1)²
Let g(x) = 3x² and h(x) = (x³ + 1)². Find g' and h' and then just plug everything into the right side of the quotient rule formula
$$\left(\frac{g}{h}\right)' = \frac{h'g - gh'}{h^2}$$
and simplify.

 

[stanton]

Now I get it. I was quite chaotic when I first saw this problem. Thank you for your advice!