Why doesn't $f(x) = x^2$ generate a counter-example to $f(-x) = -f'(x)$? According to the chain rule, the derivative of $f(-x) = -f'(x)$, since
$$
\frac{d}{dx}(f(-x)) = \frac{df(u)}{du} \frac{du}{dx} = f'(-x)*-1 = -f'(-x)
$$
where $u = -x$.  But doesn't the case of $f(x) = (x)^2$ generate a counter-example to this?
$$
f'(-x) = \frac{d}{dx}(-x)^2 = \frac{d}{dx}x^2 = 2x \ne -(2x) = -f'(-x)
$$
 A: Chain rule is
$$ \left(\frac d{dx} (f\circ g)\right) (x) = \left(\frac{df}{dx}\circ g\right)(x)\frac{dg}{dx}(x),$$
or using the $f'$ notation:
$$ (f\circ g)'(x) = (f'\circ g)(x) g'(x).$$
When we apply this with $g(x) = -x$, we see that the correct rule is
$$\frac{d}{dx} (f(-x)) = -f'(-x).$$
(This is what your derivation says, even if you stated it wrong in the title and the line before your derivation.)
We can verify this for $f(x)=x^2$ by calculating each side. On the one hand, since $f$ is even, $f\circ g= f$ so $\text{LHS}=(f\circ g)'(x)=f'(x)=2x$. On the other hand, $\text{RHS}=2(-x)\times -1=2x$, which is what I got for $\text{LHS}$.
It seems that the important error$^1$  is in saying that $-f'(-x)$ is $-2x$. In fact,
$$ f'(x) = 2x \implies f'(-x) =2(-x) = -2x \implies -f'(-x) = 2x.$$
Here are the two paths you could take
$$ f(x) \begin{cases} \xrightarrow{\text{evaluate at $-x$}} f(-x) \xrightarrow{d/dx} \frac{d}{dx} [f(-x)]\overset{\substack{Chain\\ Rule}}= -\frac{df}{dx}(-x) \\  \xrightarrow{d/dx}   \frac{df}{dx}(x) \xrightarrow{\text{evaluate at $-x$}} \frac{df}{dx}(-x) \end{cases} $$
Some of the confusion could be stemming from confusing notation. I'll list here the notations that I know: for the first I can only think of $ \frac{d}{dx}(f(-x)) = (f(-x))' =2x$
which would be the derivative of $g(y)=f(-y)=y^2$ at the point $y=x$. This is the situation in which Chain rule applies.
The other case can be written in a number of ways, $f'(-x)=\frac{df}{dx}(-x) = \left(\frac{d}{dx} f\right)(-x)  = \left.\frac{d}{dy} (y^2)\right|_{y=-x} = -2x$, these mean the function $f'(y)=2y$ evaluated at $y=-x$.  The difference here is that you differentiate $f$ first before you evaluate.
In the comments to this answer the notation $\frac{d}{d(-x)}f(-x)$ is used, but I don't recognize it. If I had to guess, I would guess that this also means $\left.\frac{d}{dy} f(y)\right|_{y=-x}=-2x$.

$^1$  as mentioned by David K and zkutch.
