$ f(x,y)=\frac{y^2-x^2}{y^2+x^2}, x_0=0, y_0=1, h=0.2,$

Step (i)

$ x=x_0+h=0+0.2=0.2 $

$ k_1 = h f(x_0,y_0)=0.2[f(0.1)]=0.2 $

$ k_2=hf(x_0+\frac{h}{2}, y_0 + \frac{k_1}{2})=0.2f(0.1,1.1)=0.19672 $

$ k_3=hf(x_0+\frac{h}{2}, y_0 + \frac{k_2}{2})=0.2f(0.1,1.0983)=0.1967 $

$ k_4=hf(x_0+h, y_0 + k_3)=0.2f(0.2,1.1967)=0.1891 $

$ k=\frac{1}{6}[k_1+2k_2+2k_3+k_4 ] $

$= \frac{1}{6}[ 0.2+ 2(0.19672) +2(0.1967)+0.19598] $

$ y=y_0+k=1+0.19598=1.19598 $

$ \hspace{20cm}$

Step (ii)

$ x_0=0.2,y_0=1.19598,h=0.2 $

$ k_1=hf(x_0,y_0)=0.2 f(0.2,1.19598)=0.18911 $

$ k_2 =hf(x_0+\frac{h}{2},y_0+\frac{k_1}{2})=0.2f(0.3,1.2905)=0.1795 $

$ k_3 =hf(x_0+\frac{h}{2},y_0+\frac{k_2}{2})=0.2f(0.3,1.28573)=0.17934 $

$ k_4 =hf(x_0+\frac{h}{2},y_0+\frac{k_3}{2})=0.2f(0.4,1.37532)=0.1688 $

$ k=\frac{1}{6}[k_1+2k_2+2k_3+k_4]=\frac{1}{6}[0.18911+2(0.1795)+2(0.17934)+0.1688]=0.1792 $

$ y=y_0+k = 1.19598+0.1792=1.3718 $