Max value

AM-GM doesn't hold for $x<0$ . You should say in your solution that the maximum of $y$ happens when $x>0$ , then use AM-MG.

x +1/x = 1/y... so value of Y will be less than one for sure. Now with negative value of x we get negative value of Y so MAX value must be 0.5 with positive sign

substitute -1 in the place of x....

Another approach:

Without loss of generality, we assume $x > 0$ since we are attaining for the maximum value of $y$ . It is impossible to get the maximum value when $x$ $\le$ $0$ , for the reason that $x^2 + 1$ $>$ $0$ for all real values of $x$ - which means if $x$ $\le$ $0$ , then the whole expression ( or the value of $y$ ) is 0 or negative in sign. Indeed, as said, by the AM - GM Inequality , we know that $\frac{{x}^2 + 1}{2}$ $\ge$ $\sqrt{{x}^2 • 1} = x$ $\Longleftrightarrow$ $\frac{x}{{x}^2 + 1} \le$ $\frac{1}{2}$ . So we're done! We have $y =$ $\frac{x}{{x}^2 + 1} \le$ $\frac{1}{2}$ . Hence, it is now clear that the maximum value of $y$ is $\frac{1}{2}$ ( when ${x}^2 = 1$ or $x = 1$ ).