How do we find the solution of the problem in interlaced form? We consider the differnetial equation $$(x-xy(x))+(y(x)+x^2)y'(x)=0$$ 
I have found the integrating factor $\mu (x,y)=\sqrt{x^2+y^2}$. 
Using this I have to find the solution of the problem in interlaced form. 
Could you give me some hints what I am supposed to do? 
I got stuck right now...  
Multiplying by the integrating factor we get $$\sqrt{x^2+y^2}(x-xy(x))dx+\sqrt{x^2+y^2}(y(x)+x^2)dy=0$$ or not? 
How could we continue? 
$$$$ 
EDIT: 
Is it maybe as follows? $$y'=F(y,x) \\ \Rightarrow (x-xy(x))+(y(x)+x^2)F(y,x)=0 \\ \Rightarrow y(x)(1-x)+x^2+x+x^2F(y,x)=0 \\ \Rightarrow y(x)=-\frac{x^2F(y,x)-x-x^2}{1-x}$$ 
 A: HINT:
$$\left(x-xy(x)\right)+\left(y(x)+x^2\right)y'(x)=0\Longleftrightarrow$$
$$\left(x^2+y(x)\right)y'(x)-y(x)x+x=0\Longleftrightarrow$$

Let $\text{Q}(x,y)=x-xy$ and $\text{T}(x,y)=x^2+y$.
This is not an exact equation, because $\frac{\partial\text{Q}(x,y)}{\partial y}=-x\ne2x=\frac{\partial\text{T}(x,y)}{\partial x}$.
Find an integrating factor $\mu(y)$ such that is exact:
$$\mu(y)\text{Q}(x,y)+\mu\cdot\frac{\text{d}y(x)}{\text{d}x}\cdot\text{T}(x,y)=0$$
This means that:
$$\frac{\partial}{\partial y}\left(\mu(y)\text{Q}(x,y)\right)=\frac{\partial}{\partial y}\left(\mu(y)\text{T}(x,y)\right)$$

$$\frac{\text{d}\mu(y)}{\text{d}y}\left(x-yx\right)-\mu(y)x=2x\mu(y)\Longleftrightarrow$$
$$\frac{\mu'(y)}{\mu(y)}=-\frac{3}{y-1}\Longleftrightarrow$$
$$\int\frac{\mu'(y)}{\mu(y)}\space\text{d}y=\int-\frac{3}{y-1}\space\text{d}y\Longleftrightarrow$$
$$\ln\left(\mu(y)\right)=-3\ln\left(y-1\right)\Longleftrightarrow$$
$$\mu(y)=\frac{1}{(y-1)^3}$$

Let $\text{P}(x,y)=-\frac{x}{(y-1)^2}$ and $\text{R}(x,y)=\frac{x^2+y}{(y-1)^3}$.
This is an exact equation, because $\frac{\partial\text{P}(x,y)}{\partial y}=\frac{2x}{y-1)^3}=\frac{\partial\text{R}(x,y)}{\partial x}$.
Define $f(x,y)$ such that $\frac{\partial f(x,y)}{\partial x}=\text{P}(x,y)$ and $\frac{\partial f(x,y)}{\partial y}=\text{R}(x,y)$:
Then, the solution will be given by $f(x,y)=\text{C}$, where $\text{C}$ is an arbitrary constant.
Integrate $\frac{\partial f(,y)}{\partial x}$ to find $f(x,y)$ and where $g(y)$ is an arbitrary function of $y$:
$$f(x,y)=\int-\frac{x}{(y-1)^2}\space\text{d}x=-\frac{x^2}{2(y-1)^2}+g(y)$$
Now, differentiate $f(x,y)$ to find $g(y)$:
$$g'(y)=\frac{y}{(y-1)^3}$$

So, in the end you've to solve:
$$-\frac{x^2}{2(y-1)^2}+\frac{1-2y}{2(y-1)^2}=\text{C}$$
A: This is how I worked this problem. I see that it is similar to Jan Eerland's in several respects, including the integrating factor of $\frac1{(1-y)^3}$. However, I think there may be enough difference to warrant posting.

Starting with the equation
$$
(x-xy)+\left(y+x^2\right)y'=0\tag{1}
$$
we want to find an integrating factor $u$ so that
$$
\frac{\partial}{\partial y}\left[u(x-xy)\right]=-xu+(x-xy)u_y\tag{2}
$$
and
$$
\frac{\partial}{\partial x}\left[u\!\left(y+x^2\right)\right]=2xu+\left(y+x^2\right)\!u_x\tag{3}
$$
are equal.
Subtracting $(2)$ and $(3)$ and dividing by $x$ yields
$$
(1-y)u_y=3u+\left(\frac yx+x\right)\!u_x\tag{4}
$$
If we make $u$ dependent only on $y$, then $u_x=0$ and $(4)$ can be solved as
$$
u=\frac1{(1-y)^3}\tag{5}
$$
Therefore, we get the exact differential equation
$$
\bbox[5px,border:2px solid #C0A000]{\mathrm{d}f=\frac{x}{(1-y)^2}\mathrm{d}x+\frac{y+x^2}{(1-y)^3}\mathrm{d}y=0}\tag{6}
$$
To solve $(6)$ for $f$, we can start by integrating up the $y$-axis, that is $x=0$, to get
$$
\begin{align}
f(0,y)-f(0,0)
&=\int_0^y\frac{t}{(1-t)^3}\,\mathrm{d}t\\
&=\frac12\frac{y^2}{(1-y)^2}\tag{7}
\end{align}
$$
Then integrate along the $x$-direction to get
$$
\begin{align}
f(x,y)-f(0,0)&=f(0,y)-f(0,0)+\int_0^x\frac{t}{(1-y)^2}\mathrm{d}t\\
&=\frac12\frac{y^2}{(1-y)^2}+\frac12\frac{x^2}{(1-y)^2}\\
&=\frac12\frac{x^2+y^2}{(1-y)^2}\tag{8}
\end{align}
$$
Therefore, the solution to $(1)$ is
$$
\bbox[5px,border:2px solid #C0A000]{\frac{x^2+y^2}{(1-y)^2}=C}\tag{9}
$$
