Separating variables, we have
[tex]\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx[/tex]
Integrate both sides.
[tex]\displaystyle \int y\,dy = \int x\,dx[/tex]
[tex]\dfrac12 y^2 = \dfrac12 x^2 + C[/tex]
Given that [tex]y(0)=-1[/tex], we find
[tex]\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12[/tex]
Then the particular solution is
[tex]\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12[/tex]
[tex]y^2 = x^2 + 1[/tex]
[tex]y = \pm\sqrt{x^2 + 1}[/tex]
and because [tex]y(0)=-1[/tex], we take the negative solution to accommodate this initial value.
[tex]\boxed{y(x) = -\sqrt{x^2+1}}[/tex]
