Answer
$$y=(2-\sqrt{x^{2}+1})^{1 / 3}$$
Work Step by Step
Given
$$x+3 y^{2} \sqrt{x^{2}+1} \frac{d y}{d x}=0, \quad y(0)=1$$
Separate
\begin{align*}
3 y^{2} \sqrt{x^{2}+1} \frac{d y}{d x} &= -x\\
\int 3 y^{2}dy&=\int \frac{- xdx}{ \sqrt{x^{2}+1} }\\
y^3&= - \sqrt{x^2+1}+C
\end{align*}
At $x= 0$, $y= 1$, then $C= 2$, and we have
$$y=(2-\sqrt{x^{2}+1})^{1 / 3}$$
