First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdot(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}=$$
$$\frac{1}{2}+\frac{2\cdot sgn(1-x^2)}{1+x^2}$$
Why did my teacher say the critical points are at $x=\pm1$? Then she wrote:
$$f_+'(1)=-\frac{1}{2}$$
$$f_-'(1)=\frac{3}{2}$$
and I'm not sure what she meant by that eather, nor by the following:
$$f'(x)=
\begin{cases}
\frac{x^2-3}{2(x^2+1)}, & |x|>1 \\
\frac{x^2+5}{2(x^2+1)}, & |x|<1
\end{cases}$$
Thank you for your time.

You have obtained that $f'(x)=\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}.$ Note that this expression doesn't exist if $x=\pm 1.$ I don't think that critical point is the most adequate word but your teacher was saying that $f'(-1)$ and $f'(1)$ don't exist. 
Now, $$f_+'(1)=\lim_{x\to 1^+}f'(x)=\lim_{x\to 1^+} \left(\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}\right)= \lim_{x\to 1^+} \left(\frac{1}{2}-\frac{2}{(1+x^2)}\right)=-\frac{1}{2},$$ where it is used that if $x>1$ then $\frac{1-x^2}{|1-x^2|}=-1.$ The value of $f_-'(1)$ can be obtained in a similar way.
Finally, note that if $|x|>1$ then $\frac{1-x^2}{|1-x^2|}=-1$ and if $|x|<1$ then $\frac{1-x^2}{|1-x^2|}=1.$ So, you have two different expressions for $f'(x):$ one if $|x|>1$ and other if $|x|<1.$ 
