ODE solution how to look at $f(x)$ as if it is a function of $x$ $ y=\left(y^{3}+2x\right)y' $
$y(0)=1$
I've been told to look at $y$ as a function of $x$. It is possible around $x=0$ because $y(0)=1$ and not $0$.
But I don't have an idea what to do next.
Thanks for the help. You guys are great!
 A: if you rewrite your differential equations as $$ \frac{dx}{dy} - \frac{2x}{y} = y^2$$ you can find the integrating factor $a$ by requiring $$\frac{da}{dy} = -\frac{2a}{y} \to a = \frac 1{y^2}$$ so that $$1= \frac 1{y^2} \frac{dx}{dy} - \frac{2x}{y^3} = \frac{d}{dy}\left(\frac x{y^2}\right) $$ inetgrating with respect to $y,$  we get $$ y-1 =\frac{x}{y^2} \to x = y^2(y-1). $$
A: Just rewrite the equation as follows:
$$
y = (y^3 + 2 x) \frac{dy}{dx} \quad
\Longrightarrow
\quad
\frac{dx}{dy} y = y^3 + 2 x.
$$
Now solve first the homogeneous equation (by separation of variables)
$$
\frac{dx}{dy} y = 2 x.
$$
And then use the variation of arbitrary constant to get the desired solution.
A: For an equation of the form
$$
\partial_x{F(x,y)}\mathbf{d}x+\partial_y{F(x,y)}\mathbf{d}y=\mathbf{d}F=0
$$
the solution is given implicitely by $F(x,y)=$const.
Unfortunately your equation 
$$
y\mathbf{d}x-(y^3+2x)\mathbf{d}y=0
$$
isn't of this form. But by multiplying with a kernel, in our case $y^{-3}$, we get
$$
y^{-2}\mathbf{d}x-(1+2xy^{-3})\mathbf{d}y=\mathbf{d}F=0\,,
$$
for $F=xy^{-2}-y$. For the given initial conditions we get a solution $xy^{-2}-y=-1$ ie $x=y^2(y-1)$. This can then be inverted (which is a bit of a mess).
NB: For this method, finding a suitable kernel is the hard part. It is usually a good idea to try powers of $x$ and $y$.
