Answer
$\displaystyle y = \sqrt[3] {2-\sqrt{x^2+1}} = (2-(x^2+1)^{1/2})^{1/3}$
Work Step by Step
$\displaystyle x+3y^2\sqrt{x^2+1}\cdot\frac{dy}{dx}=0$
Seperation of Variables
$\displaystyle 3y^2\sqrt{x^2+1}\cdot\frac{dy}{dx} = -x$
$\displaystyle 3y^2\frac{dy}{dx} = -\frac{x}{\sqrt{x^2+1}}$
$\displaystyle \int3y^2dy = -\int\frac{x}{\sqrt{x^2+1}}dx$
Integrate each side. The right side uses $u$-substitution where $u = x^2+1$
$\displaystyle y^3 = -\int u^{-1/2}\cdot\frac{1}{2}du\qquad\qquad u=x^2+1\quad\rightarrow\quad \frac{1}{2}du = xdx$
$\displaystyle y^3 = -\frac{1}{2}(\frac{u^{1/2}}{1/2})+C$
$\displaystyle y^3 = -\frac{1}{2}(2u^{1/2})+C$
$\displaystyle y^3 = -u^{1/2}+C$
$\displaystyle y^3 = -(x^2+1)+C = -\sqrt{x^2+1}+C$
Plug in the initial condition to solve for $C$
$\displaystyle (1)^3 = -\sqrt{(0)^2+1} +C=-\sqrt{1}+C = -1+C$
$2 = C$
Plug $C$ back into the equation and isolate $y$
$\displaystyle y^3 = -\sqrt{x^2+1}+2$
$\displaystyle y = \sqrt[3] {-\sqrt{x^2+1}+2} = (-(x^2+1)^{1/2}+2)^{1/3}$
