$f(x^2)$ even or odd I've been working on the following example:
Is the following even, odd or neither: $f_{0}(x^2)$, where $f_{0}(x)$ can be any unknown function
I've tried the following:

1) for example I assume $$f_{0}(x^2)=x^3$$ Then: $$f_{0}(x^2)=x^2 \cdot x$$ $$f_{0}(x^2)=x^2 \sqrt{x^2}$$
$$f_{0}(x)=x \sqrt{x}$$
Now I take $f_{0}(-x)$ which is: $$f_{0}(-x)=-x^{1.5}$$
This is neither $f_{0}(x)$(would be even) nor $-f_{0}(x)$(would be odd) so it is neither even nor odd. Is this true?

2) My second attempt is: 
$f_{0}((-x)^2)=f_{0}(x^2)$ which shows that it is even.

I got 2 opposite results. Which of my attempts is true?
 A: I think you are getting confused by $f(x^2)$. $f$ is a function $f:x\longmapsto f(x)$. 
Consider $g:x\longmapsto g(x)=f(x^2)$.
$\forall x\in\mathbb{R},\ g(-x)=g(x)$. Therefore $g$ is even. So is the function $x\longmapsto f(x^2)$ of course.
In your first attempt you did not compare $f((-x)^2)$ and $f(x^2)$.
A: Let's just prove a more general proposition:
Given two functions $f,g$ with $g$ even, the composition $f\circ g$ is even, with no conditions on $f$.
Proof: $\forall x\quad f\circ g(-x)=f(g(-x))=f(g(x))=f\circ g(x)$   QED
A: Let $g(x)=f(x^2)$ then
$$g(-x)=f((-x)^2)=f(x^2)=g(x)$$
so it is even
A: In your first attempt, you attempt to "solve" for a function $f_0$ such that $f_0(x^2) = x^3$.    You come to the conclusion that if we define $f_0(x) = x^{1.5}$, then we'd have $f_0(x^2) = x^3$. 
This is true if you only consider $x \geq 0$, but it is not true for $f_0$ over all of $\Bbb R$.  In your third line, you make the substitution $x = \sqrt{x^2}$, which only applies when $x \geq 0$  (otherwise, we'd have $x = -\sqrt {x^2}$).  As a result, the conclusion you reached was invalid.
