Looks like the equation is
[tex]yy'+x=\sqrt{x^2+y^2}[/tex]
Substitute [tex]u(x)=x^2+y(x)^2[/tex], so that [tex]u'=2x+2yy'[/tex]. Then the equation is the same as
[tex]\dfrac{u'-2x}2+x=\sqrt u\implies u'=2\sqrt u\implies\dfrac{\mathrm du}{2\sqrt u}=\mathrm dx[/tex]
Integrate both sides to get
[tex]\sqrt u=C\implies\sqrt{x^2+y^2}=C[/tex]
Given that [tex]y(1)=\sqrt8[/tex], we have
[tex]\sqrt{1^2+(\sqrt8)^2}=C\implies C=3[/tex]
so the solution is
[tex]\sqrt{x^2+y^2}=3[/tex]
