This ODE is separable, so you can write
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2}{x^3}\iff\dfrac{\mathrm dy}{y^2}=\dfrac{\mathrm dx}{x^3}[/tex]
Integrating both sides gives
[tex]-\dfrac1y=-\dfrac1{2x^2}+C[/tex]
and given the initial condition [tex]y(1)=1[/tex], you have
[tex]-\dfrac11=-\dfrac1{2(1)^2}+C\implies C=-\dfrac12[/tex]
so that the solution is
[tex]-\dfrac1y=-\dfrac1{2x^2}-\dfrac12\implies y=\dfrac1{\frac1{2x^2}+\frac12}=\dfrac{2x^2}{1+x^2}[/tex]