If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$ If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$
My Attempt:
$$p^3-p(x^2+xy+y^3)+x^2y+xy^2=0$$
$$p^3-p(x+y)^2+pxy+xy(x+y)=0$$
$$p(p^2-(x+y)^2)+xy(p+(x+y))=0$$
$$p\cdot (p+(x+y))\cdot (p-(x+y))+xy(p+(x+y))=0$$
$$(p+(x+y))\cdot (p(p-(x+y))+xy)=0$$
I don't get what to do after this.
 A: Your second factor factors further. You have 
$$(p+x+y)(p-x)(p-y)=0$$
At any $(x,y)$ on a solution, either $p=-x-y$, $p=x$, or $p=y$.
That is: $$\frac{dy}{dx}=-x-y\implies y=ke^{-x}-x+1$$
Or $$\frac{dy}{dx}=x\implies y=\frac{1}{2}x^2+C$$
Or $$\frac{dy}{dx}=y\implies y=k e^x$$

There are also solutions that combine these forms, where a piecewise definition lets you transfer from one to the other. For example, 
$$y=\begin{cases}
\frac{1}{2}x^2&x<0\\
0e^x&x\geq 0
\end{cases}$$
or 
$$y=\begin{cases}
-e^x&x<-W(1)\\
\frac{1}{2}x^2-W(1)-\frac12W(1)^2&x\geq -W(1)
\end{cases}$$
where $W$ is the Lambert function. (What's important is that $e^{-W(1)}=W(1)$.)
This solution is continuous and differentiable at $W(1)$ where the transition takes place.

I don't think it would be too hard to catalog all such chimera solutions, built form the three primary solutions. You would need to solve for what conditions allow any two of the primary solutions to meet continuously and differentiably.
A: $$(p+(x+y))\cdot (p(p-(x+y))+xy)=0$$
Solve the first DE:
$$p+(x+y)=0$$
$$(ye^x)'=-xe^x$$
Integrate.
$$ye^x=-xe^x+e^x+c$$
$$\implies y=-x+1+ce^{-x}$$

The second DE can be factorized as:
$$p^2-p(x+y)+xy=0$$
$$(p-x)(p-y)=0$$
You can easily solve both equations.
$$y'=x \implies y=\dfrac {x^2}2+c_2 $$
$$y'=y \implies (ye^{-x})'=0 \implies y=c_1e^x$$
A: From here you can interpret two cases,
Case I: $p=-(x+y)$
$$\dfrac{dy}{dx}+y=-x$$
$$IF(integrating\space factor)={e^x}$$
$$ye^{x}=\int -xe^{x}$$
$$y=1-x+Ce^{-x}$$
Now Case II: $p(p-(x+y))+xy=0$
$$p^2-(x+y)p+xy=0$$
$$\therefore p=x\space or\space p=y\space (x\space and\space y\space\ are\space roots\space of\space this\space equation)$$
$$\dfrac{dy}{dx}=x\Rightarrow y=\dfrac{x^2}{2}+C$$
$$\dfrac{dy}{dx}=y\Rightarrow y=Ce^x $$
A: We get three equations: $p=x,y-x-y$
We get three solutions: $$y=x^2/2+C, y=Ce^x, y=1-x+Ce^{-x}$$
The total solution of this first order ODE is
$$(x^2/2-C)(ye^{-x}-C)[(y-1+x)e^{x}-C]=0,$$
with only one contant of integration: $C$.
