A separable first-order equation

Separation. Multiply both sides by $(1+2y)$:
$$(1+2y)\frac{dy}{dx} = x \implies (1+2y)dy = x\,dx.$$

Integration.
$$\int (1+2y)\,dy = \int x\,dx \implies y + y^2 = \frac{x^2}{2} + C.$$

Initial condition. With $y(0)=0$:
$$0 + 0 = 0 + C \implies C=0.$$

Implicit solution.
$$y^2 + y - \frac{x^2}{2} = 0.$$
Solving for $y$:
$$y = \frac{-1 \pm \sqrt{1+2x^2}}{2}.$$
Since $y(0)=0$, we take the $+$ branch:
$$y(x) = \frac{-1 + \sqrt{1+2x^2}}{2}.$$