solution of a first order differential equation with involving the inverse of a function which trick should i use to solve the nonlinear differential equation 
$$ y(y'(x))=x $$
A composition of a function with its first derivative
i guess that for some A and B has the solution $ y(x) = Ax^{2}+bx+C $
are there another solutions ?, the initial condition is not specified
 A: Well, we know that:
$$\text{y}\left(\text{y}^{-1}\left(x\right)\right)=x\tag1$$
Where $\text{y}^{-1}\left(x\right)$ is the inverse of the function $\text{y}\left(x\right)$.
So, using your problem we can write:
$$\text{y}'\left(x\right)=\text{y}^{-1}\left(x\right)\tag2$$
Note that a power of $x$ fits the bill for the differential equation, given in $(2)$. So, let's set:
$$\text{y}\left(x\right)=\text{A}x^\text{r}\tag3$$
We see that:
$$\text{y}'\left(x\right)=\text{r}\text{A}x^{\text{r}-1}\tag4$$
Now the inverse of the function is given by:
$$\text{y}^{-1}\left(x\right)=\left(\frac{x}{\text{A}}\right)^\frac{1}{\text{r}}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\tag5$$
So, we need to look at:
$$\text{r}\text{A}x^{\text{r}-1}=\left(\frac{1}{\text{A}}\right)^\frac{1}{\text{r}}\cdot x^\frac{1}{\text{r}}\space\Longleftrightarrow\space x^{\text{r}-1-\frac{1}{\text{r}}}=\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}\tag6$$

Now, to finish note that the RHS is a constant, so the LHS is a constant which means that $\text{r}-1-\frac{1}{\text{r}}=0$, which means that $\text{r}=\frac{1\pm\sqrt{5}}{2}$. So the LHS gives $x^0=1$, which means that $\frac{1}{\text{r}}\left(\frac{1}{\text{A}}\right)^{1+\frac{1}{\text{r}}}=1$, and that gives $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$.

Concluding, we see that this is indeed true for $\text{r}=\frac{1+\sqrt{5}}{2}$ and $\text{A}=\left(\frac{2}{1+\sqrt{5}}\right)^\frac{2}{1+\sqrt{5}}$:
$$\text{y}\left(\text{y}'\left(x\right)\right)=\text{A}\left(\text{r}\text{A}x^{\text{r}-1}\right)^\text{r}=x\tag7$$
Note that when $\text{r}=\frac{1-\sqrt{5}}{2}$ there is no solution.
