3. Problem: Find all real numbers $x$ and $y$ such that
$$
\begin{aligned}
x^{2}+y^{2} & =2, \\
\frac{x^{2}}{2-y}+\frac{y^{2}}{2-x} & =2 .
\end{aligned}
$$
Solution: (Viet Hoang)
From the second equation, after a few steps of algebraic manipulation, one has
$$
\begin{aligned}
x^{2}(2-x)+y^{2}(2-y) & =2(2-x)(2-y) \\
2\left(x^{2}+y^{2}\right)-(x+y)\left(x^{2}+y^{2}-x y\right) & =8-4(x+y)+2 x y \\
4-(x+y)(2-x y) & =8-4(x+y)+2 x y \\
-2(x+y)+x y(x+y) & =4-4(x+y)+2 x y \\
2(x+y)+x y(x+y) & =4+2 x y \\
(x+y-2)(2+x y) & =0
\end{aligned}
$$

In the case $x y=-2$, substituting this into the first equation, we have $x^{2}+x y+y^{2}=0$. However, this means
$$
\left(x+\frac{y}{2}\right)^{2}+\frac{3 y^{2}}{4}=0 \Longrightarrow x=y=0
$$

Substituting these values into the second equation, it yields a contradiction. For the second case, using the substitution $y=2-x$, we get
$$
\begin{array}{r}
x^{2}+y^{2}=2, \\
x+y=2 .
\end{array}
$$

From here we get: $(x-1)^{2}+(y-1)^{2}=\left(x^{2}+y^{2}\right)-2(x+y)+2=2-4+2=0$. Therefore $x=y=1$.