$4 \cdot 115$ Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+\frac{2 x y}{x+y}=1, \\
\sqrt{x+y}=x^{2}-y .
\end{array}\right.
$$
[Solution] From $x^{2}+y^{2}+\frac{2 x y}{x+y}=1$, we get
$$
(x+y-1)\left(x^{2}+y^{2}+x+y\right)=0
$$

If $x^{2}+y^{2}+x+y=0$, then
$$
x+y=-\left(x^{2}+y^{2}\right) \leqslant 0 .
$$

From the original equation, we know $x+y \geqslant 0$.
Therefore, $x+y=0$, but for the original equation, $x+y=0$ cannot be the denominator. Thus,
$$
x^{2}+y^{2}+x+y \neq 0 \text {. }
$$

Hence,
$$
x+y-1=0 \text {. }
$$

Solving the system of equations $\left\{\begin{array}{l}x+y=1, \\ x^{2}-y=1 .\end{array}\right.$
we get $\left\{\begin{array}{l}x_{1}=1, \\ y_{1}=0 ;\end{array}\left\{\begin{array}{l}x_{2}=-2, \\ y_{2}=3 .\end{array}\right.\right.$