How to solve this nonstandard system of equations? How to solve this system of equations
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1+(1-y)\sqrt{x}.
\end{cases}$$
I see $(0,1)$ is a root.
 A: Solution.
First way
From the first equation, we have 
$$\begin{cases}
2x^2\leqslant 1,\\
y^2 \leqslant 1
\end{cases}
\Leftrightarrow 
\begin{cases}
-\dfrac{1}{\sqrt{2}} \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
Then, the conditions of $x$ and $y$ are
$$\begin{cases}
0 \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
We have $x^2 + y^2 = 1-x^2$. Therefore $x^2 + y^2 \leqslant 1$.
Another way,
$$1-x^2 =  y \sqrt{1-x^2} -(1-y)\sqrt{x} \leqslant  y \sqrt{1-x^2}.$$
Because
$$ y \sqrt{1-x^2} \leqslant \dfrac{y^2 + 1 - x^2}{2}.$$
Implies
$$1-x^2  \leqslant \dfrac{y^2 + 1 - x^2}{2}  \Leftrightarrow x^2 + y^2 \geqslant 1 .$$
From $x^2 + y^2 \leqslant 1$ and $x^2 + y^2 \geqslant  1$, we have $x^2 + y^2 = 1.$
Solve 
$$\begin{cases}
x^2 + y^2 = 1,\\
2x^2 + y^2 = 1,\\
0 \leqslant  x \leqslant 1,\\
- 1 \leqslant y \leqslant \dfrac{1}{\sqrt{2}}
\end{cases} \Leftrightarrow \begin{cases}
x = 0,\\
y = 1.\end{cases}$$
Second way.
We have $2x^2 + y^2 = 1$, therefore $y=\sqrt{1 - 2x^2}$ or $y=-\sqrt{1 - 2x^2}.$
First case, $y=\sqrt{1 - 2x^2}$, subtitution the second equation, we get
$$x^2 + \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2}=1+(1-\sqrt{1 - 2x^2})\sqrt{x}.$$
equavalent to
$$1 - x^2 - \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2} + (1-\sqrt{1 - 2x^2})\sqrt{x}=0$$
or
$$\sqrt{1-x^2} (\sqrt{1-x^2} - \sqrt{1 - 2x^2}) + (1-\sqrt{1 - 2x^2})\sqrt{x}=0.$$
This is equavalent to
$$\dfrac{\sqrt{1-x^2}\cdot(1-x^2-1+2x^2)}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}}+\dfrac{(1-1+2x^2)\sqrt{x}}{1+\sqrt{1 - 2x^2}} = 0$$
or 
$$x^2\left (\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}\right )=0.$$
It is easy to see that
$$\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}> 0$$
Thus $x = 0$.
Second cases.  $y=-\sqrt{1 - 2x^2}.$
 We can check that
$$x^2 + y \sqrt{1-x^2} \leqslant \dfrac{1}{2}$$
and
$$1+(1-y)\sqrt{x} >1.$$
In this case, the given system of equations has no solution. 
A: We solve the second system for $y$:
$$y=\frac{-x^2+\sqrt{x}+1}{\sqrt{x}+\sqrt{1-x^2}}$$
and substitute into the first equation and solve for $x$. It is then seen that $(0,1)$ is the only real-valued solution. Computer analysis finds complex solutions where $x$ is the root of a certain $16$ degree polynomial.
Edit: the precise polynomial is $$ x^{16}+12 x^{15}+30 x^{14}-96 x^{13}-79 x^{12}+360 x^{11}+70 x^{10}-804 x^9-92 x^8+972 x^7+230 x^6-600 x^5-207 x^4+192 x^3+62 x^2-36 x+1 =0$$
A: (More a comment.) If we allow the non-principal square root,
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1\color{red}{\pm} (1-y)\sqrt{x}
\end{cases}$$
the $+$ case has one real solution, but the $-$ case has three real solutions: $(x,y)=(0,1)$ and two which surprisingly are roots of $11$-deg equations. Using,
$$x=+\sqrt{\frac{1-y^2}{2}}$$
and with $y$ as two appropriate real roots of,
$$\small49 - 301 y + 327 y^2 - 51 y^3 - 1038 y^4 + 1334 y^5 - 1066 y^6 + 850 y^7 - 259 y^8 + 279 y^9 + 3 y^{10} + y^{11}=0$$
namely $y_1 \approx -0.619107$, and $y_2 \approx 0.939251$.
