Answer
$y(1)=0.512$
Work Step by Step
We have
$$y'=-\frac{2xy}{1+x^2}$$
Setting $h=0.1$ in equation $y'=-\frac{2xy}{1+x^2}$
$$y_{n+1}=y_n+0.1(-\frac{2xy}{1+x^2})$$
Hence,
$y_1=y_0+0.1(-\frac{2xy}{1+x^2})=1$
$y_2=y_1+0.1(-\frac{2xy}{1+x^2})=0.98$
$y_3=y_2+0.1(-\frac{2xy}{1+x^2})=0.942$
$y_4=y_3+0.1(-\frac{2xy}{1+x^2})=0.891$
$y_5=y_4+0.1(-\frac{2xy}{1+x^2})=0.829$
$y_6=y_5+0.1(-\frac{2xy}{1+x^2})=0.763$
$y_7=y_6+0.1(-\frac{2xy}{1+x^2})=0696$
$y_8=y_7+0.1(-\frac{2xy}{1+x^2})=0.61$
$y_9=y_8+0.1(-\frac{2xy}{1+x^2})=0.569$
$y_{10}=y_9+0.1(-\frac{2xy}{1+x^2})=0.512$
If we increase the step size to $h = 0.1$, the corresponding approximation to $y(1)$ becomes $y_{10}=0.512$
