Solve the initial value problem $y' = \frac{xy^3}{\sqrt{1+x^2}}$ with $y(0)=-1$ 
Solve the initial value problem $y' = \frac{xy^3}{\sqrt{1+x^2}}$ with
  $y(0)=-1$

I'd like to know if my solution is correct. To keep it short, just assume that all integration steps I did are right.
The differential equation isn't in its standard form $y'+P(x)y=Q(x)$, so we bring it to that form:
$$y'-\frac{x}{\sqrt{1+x^2}}y^3=0$$
And at this step is my first doubt; is it really the standard form? It doesn't have to be $y$ instead of $y^3$?
Continuing, the integration factor is $e^{\int{P(x)dx}}=e^{\int{\frac{x}{\sqrt{1+x^2}}}dx}=e^{\sqrt{1+x^2}}$, in next step we will multiply the differential equation with $\sqrt{1+x^2}$.
$$\Rightarrow  \frac{dy}{dx}-\frac{x}{\sqrt{1+x^2}}y^3=0 \Leftrightarrow \sqrt{1+x^2} \cdot\frac{dy}{dx}- \sqrt{1+x^2} \cdot \frac{x}{\sqrt{1+x^2}}y^3=0 \Leftrightarrow \frac{d}{dx}(\sqrt{1+x^2} \cdot y)-x \cdot y^3=0 \Leftrightarrow$$ 
$$\Leftrightarrow \int{\frac{d}{dx}(\sqrt{1+x^2} \cdot y)-x \cdot y^3} = \int{0} \text{ }dx \Leftrightarrow \sqrt{1+x^2} \cdot y = c \Leftrightarrow y = \frac{1}{\sqrt{1+x^2}}c$$
We know that $y(0)=-1 \Leftrightarrow -1=\frac{1}{\sqrt{1+0^2}}c \Leftrightarrow c=-1$
$\Rightarrow y = -\frac{1}{\sqrt{1+x^2}}$

You are making it yourself way too hard here.
We have
$$\frac{dy}{dx} = \frac{xy^3}{\sqrt{1+x^2}}$$
or equivalently:
$$\frac{dy}{y^3} = \frac{xdx}{\sqrt{1+x^2}}$$
Now integrate both sides, to obtain:
$$\frac{-1}{2y^2} = \sqrt{1+x^2} + c$$
Now solve for $y$ and fill in the initial values.
