Finding dy/dx with two elements? I don't understand the problem. How can I find $\frac{dy}{dx}$, given that $y = \frac{u^2 - 1}{u^2 + 1}$ and $u = \sqrt[3]{x^2 + 2}$?
For the next step, the book says:
$\frac{dy}{du} = \frac{4u}{(u^2+1)^2}$.
Am I supposed to place the value of $u$ into the $y$ fraction?
If I put it into the numerator, it won't solve for 4u. Is the book wrong?
Also, where does du come from? It says dx before.
 A: You need to use the Chain Rule, which states that $$\dfrac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$$
In particular, if $y(u(x)) = \frac{u(x)^2 - 1}{u(x)^2 + 1}$ and $u(x) = \sqrt[3]{x^2 + 2}$, then $$\dfrac{d}{dx} y(u(x)) = y'(u(x)) u'(x)$$
You can differentiate $u(x)$ yourself to get $u'(x)$; and $$y(r) = \frac{r^2-1}{r^2+1} \Rightarrow y'(r) = \frac{2r(r^2+1)-2r(r^2-1)}{(r^2+1)^2}$$
by the quotient rule, so you can let $r=u(x)$ to obtain the expression for $y'(u(x))$.
This process is often abbreviated as $$\frac{dy}{dx} \frac{dx}{dz} = \frac{dy}{dz}$$
but do remember that these are not fractions, but shorthands for derivatives with respect to variables. It is not simply a matter of "cancelling out $dx$", and if you try that with three variables you will get the wrong answer!
A: You need to apply this fact:

The Chain Rule: Let $z(y)$ and $y(x)$ be differentiable functions. Then $z(y(x))$ is
  differentiable with respect to $x$ and
  $$\frac{dz}{dy}\cdot\frac{dy}{dx}=\frac{dz}{dx}$$

That should solve all your problems, since you can find $\frac{du}{dx}$ pretty easily. Notationally, it's generally accepted that you should use substitution to eliminate all instances of $y$ in the resulting expression (or in your case, $u$).
A: What the book is doing is it is going to take $\frac{\text{d}y}{\text{d}u}$ and multiply it by $\frac{\text{d}u}{\text{d}x}$; atleast that's what I would expect. As it stated
$$\frac{\text{d}y}{\text{d}u}=\frac{4u}{(u^2+1)^2}$$
Then we also have
$$\frac{\text{d}u}{\text{d}x}=\frac{2x}{3\sqrt[3]{(x^2+2)^2}}$$
To end up with
$$\frac{\text{d}y}{\text{d}x}=\frac{8xu}{3(u^2+1)^2\sqrt[3]{(x^2+2)^2}}$$
and substitute in $u$ in terms of $x$. 
Similarily, you could just use the chain rule in a slightly different fashion keeping in mind that $u$ is a function of $x$. If we differentiate $y$ what we get is
$$\frac{\text{d}y}{\text{d}x}=\frac{2u\cdot u'(u^2+1)-2u\cdot u'(u^2-1)}{(u^2+1)^2}$$
and if you put $u$ in terms of $x$ in both of these and simplify we should find that we got the derivative of $y$ with respect to $x$ to be the same.
A: You have two functions:
$$
y=f(u)=\frac{u^2-1}{u^2+1} \qquad u=g(x)=\sqrt[3]{x^2+1}
$$
so $y$ is a composite function:
$$
y=f(g(x))=\frac{\left(\sqrt[3]{x^2+1}\right)^2-1}{\left(\sqrt[3]{x^2+1}\right)^2+1}
$$
Maybe that, in this form, you know how find the derivative using the chain rules:
$$
\frac{dy}{dx}=f'(g(x))g'(x)
$$
Anyway, your book  gives you the derivative 
$$\frac{dy}{du}=\frac{df}{du}=f'(u)=f'(g(x))=\frac{4u}{(u^2+1)^2}=\frac{4\sqrt[3]{x^2+1}}{\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}$$
so you have to calculate 
$$
g'(x)=\frac{d}{dx}\sqrt[3]{x^2+1}=\frac{2x}{3\sqrt[3]{(x^2+1)^2}}
$$
and multiply to complete the chain rule.
$$
\frac{dy}{dx}=\frac{4\sqrt[3]{x^2+1}}{\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}\frac{2x}{3\sqrt[3]{(x^2+1)^2}}=
\frac{8x}{3\sqrt[3]{x^2+1}\left(\left(\sqrt[3]{x^2+1}\right)^2+1\right)^2}
$$

Deriving directly the function $y=f(g(x))$, using the quotient and the chain rules you can find:
$$
y'=\frac{8}{3}x\frac{1}{\sqrt[3]{x^2+1}}\frac{1}{\left(\left(\sqrt[3]{x^2+1} \right)^2+1\right)^2}
$$
note that:
$$
\frac{8}{3}x\frac{1}{\sqrt[3]{x^2+1}}=4\sqrt[3]{x^2+1}\left(\frac{2}{3}x\frac{1}{\sqrt[3]{(x^2+1)^2}}\right)=4u(x)u'(x)
$$
