Prove that the derivative of an even differentiable function is odd, and the derivative of an odd is even.

Continuing from your last line, $$ \lim_{h\to 0} -\frac {f(x-h)-f(x)}{h}=\lim_{h\to 0} \frac {f(x-h)-f(x)}{-h}=f'(x) $$

That completes the proof for $f$ an odd function.

The analogous approach will probably work for $f(x)$ an even function.

$$ \lim_{h\to 0} \frac {f(-x+h)-f(-x)}{h}=\lim_{h\to 0} \frac {f(x-h)-f(x)}{h}=-\lim_{h\to 0} \frac {f(x-h)-f(x)}{-h}=-f'(x) $$

If $f(x)$ is odd then,$$f'(x)=\frac{d(f(x))}{dx}=\frac{d(-f(-x))}{dx}=-\frac{d(f(-x))}{dx}=-(-f'(-x))=f'(-x)$$

Here, have used $\frac{d(f(-x))}{dx}=-f'(-x)$ (using chain rule)

You can follow similar approach for even $f(x)$