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.

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:
$$
\frac{8}{3}x\frac{1}{\sqrt[3]{x^2+1}}=4\sqrt[3]{x^2+1}\left(\frac{2}{3}x\frac{1}{\sqrt[3]{(x^2+1)^2}}\right)=4u(x)u'(x)
$$
