Solve $dy/dx=x^2+xy+y^2$. Wolfram say's that the answer is $y=\dfrac{1}{c-x}-\dfrac{x}{2}$. But i don't know how to get there. I try this, $dy/dx=(x+y)^2-xy$. And substitute $y=vx$, but i don't get anywhere. Any help?
 A: $y=\dfrac{1}{c-x}-\dfrac{x}{2}\quad$ is not solution of $\quad \dfrac{dy}{dx}=x^2+xy+y^2\tag 1$.
$y=\dfrac{1}{c-x}-\dfrac{x}{2}\quad$ is solution of $\quad \dfrac{dy}{dx}=\frac14 x^2+xy+y^2-\frac12\tag 2$.
Both ODEs are of Riccati kind. The change of function is
$$y(x)=-\frac{u'(x)}{u(x)} \quad\implies\quad y'=-\frac{u''}{u}+\frac{(u')^2}{u^2}$$
Case of ODE $(2)$ :
$$\dfrac{dy}{dx}=\frac14 x^2+xy+y^2-\frac12=-\frac{u''}{u}+\frac{(u')^2}{u^2}=\frac14 x^2+x\left(-\frac{u'}{u}\right)+\left(-\frac{u'}{u}\right)^2-\frac12$$
$$u''-xu'+\left(\frac14 x^2-\frac12\right)u=0$$
This is a linear second order ODE. Solving it leads to :
$$u=(c_1+c_2 x)\exp(\frac14 x^2)$$
$$y=-\frac{u'}{u}=-\frac{c_1x+2c_2+c_2x^2}{2(c_1+c_2x)}$$
And with $c=-\frac{c_1}{c_2}$ we get to the expected result :
$$y=\dfrac{1}{c-x}-\frac{x}{2}$$
Case of ODE $(1)$ :
$$\dfrac{dy}{dx}=x^2+xy+y^2=-\frac{u''}{u}+\frac{(u')^2}{u^2}=x^2+x\left(-\frac{u'}{u}\right)+\left(-\frac{u'}{u}\right)^2$$
$$u''-xu'+x^2u=0$$
This linear second order ODE is solvable involving confluent hypergeometric functions. I suppose that you are not interested for the complicated final analytic solution $y(x)$.
