System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$ 
Solve over reals:
$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$

I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:
$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$
Here, there are two cases: all three variables are greater than $1$ or only one. 
If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.
How can I prove there are no solutions when only variable, say $x$, is greater than $1$?

I will prove by contradiction that there exists no solutions for $x > 1$ and $y,z < 1$. Suppose $(x,y,z)$ is such a solution.
Iteration #1: First, note that:
$$
z^3 = x^2 - x = x(x - 1) > 0 \implies z > 0
$$
Thus, $0 < z < 1$. We see that:
$$
y^3 = z^2 - z = z(z - 1)
$$
For $0 < z < 1$, $-\frac{1}{4} < z(z - 1) < 0$, so $-2^{-\frac{2}{3}} < y < 0$. For this range of values of $y$, since $y(y - 1)$ is decreasing for $y < 0$, we have that:
$$
x^3 = y^2 - y < 2^{-\frac{4}{3}} + 2^{-\frac{2}{3}} = 1.0268108 \implies x < 1.00885823
$$
Iteration #2: Applying this inequality to the third equation again, we have:
$$
z^3 = x(x - 1) < 1.00885823(1.00885823 - 1) = 0.0089367 \implies z < 0.20752
$$
For $0 < z < 0.20752$, we have that:
$$
0.20752(0.20752 - 1)  < y^3 < 0 \implies -0.54787 < y < 0
$$
Once again, we have:
$$
x^3 = y(y - 1) < -0.54787 \left(-0.54787 - 1\right) < 1 \implies x < 1
$$
This contradicts that $x > 1$. Therefore, no non-trivial solutions exist.
