Solving the equation $ f^{-1}(x)=f(x)$ I attempted to solve the equation given in the title for the function;

$$f: \mathbb R_{++} \to\mathbb R_{++}; \quad f(x)=x^2(x+2)$$

I understand that the problem is equivalent to solving $f(f(x))=x$
but since this seemed like too much work, I had a look at the solution and it stated that;

$$ f^{-1}(x)=f(x) \Longrightarrow  f(x)=x$$

I don't understand why this is the case. Can someone please explain this?

Thanks
Although $f(f(x))=x\implies f(x)=x$ is not always true, it is true if $f$ is an increasing function (even if only weakly increasing).  

Suppose that $f(x)=y$ and $f(y)=x$. If $x\le y$, then $f(x)=y\le f(y)=x$, and vice-versa.
So either way, both $x\le y$ and $y\le x$ are true, and thus $x=y$ and $f(x)=x$.

So your book is not off the rails; this is a valid approach to the problem.
Answering this problem is simple.

f(x) = x^2(x+2)

Thus f(x) = a(x)b(x) where a(x) = x^2 and b(x) = x+2

The product rule tells us that:

f'(x) = 3x^2 + 4x

Thus solve for f(x) = f(x)':

x^2(x+2) = 3x^2 + 4x

x(x^2 -x - 4) = 0

Therefore x = 0; x = (1±sqrt[17])/2