How do we show that this modular function is one-to-one? $$f(x) = {(x^2+5)}\bmod{9}$$
$${(x^2 + 5)} \bmod {9} = (x^2 + 5)\bmod 9$$
$$(x^2 + 5) = (x^2 + 5)$$
$$x^2 = x^2$$
$$x = x$$
Is this the correct way to do this?
I have no idea how to manipulate the terms.

The function is not one to one. To show this, you need to show that there exist $a$ and $b$ such that $a\not\equiv b\pmod{9}$ but $a^2+5\equiv b^2+5\pmod{9}$. It should not take long to find such a pair $(a,b)$: there are several. 
Remark: You can operate partly by analogy. The reason the function $x^2+5$ is not one to one in this setting is related to the reason $x^2+5$ is not one to one on the reals.  
