10. (20 points) Let positive numbers $x, y$ satisfy $x^{3}+y^{3}=$ $x-y$. Find the maximum value of the real number $\lambda$ such that $x^{2}+\lambda y^{2} \leqslant 1$ always holds.
10. Given positive numbers $x, y$ satisfying $x^{3}+y^{3}=x-y$, we know $x>y>0$.
Let $t=\frac{x}{y}>1$. Then
$$
\begin{array}{l}
x^{2}+\lambda y^{2} \leqslant 1 \Leftrightarrow x^{2}+\lambda y^{2} \leqslant \frac{x^{3}+y^{3}}{x-y} \\
\Leftrightarrow \lambda y^{2} \leqslant \frac{x^{3}+y^{3}}{x-y}-x^{2}=\frac{x^{2} y+y^{3}}{x-y} \\
\Leftrightarrow \lambda \leqslant \frac{x^{2} y+y^{3}}{(x-y) y^{2}} \\
\Leftrightarrow \lambda \leqslant \frac{x^{2}+y^{2}}{x y-y^{2}}=\frac{t^{2}+1}{t-1} .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
f(t)=\frac{t^{2}+1}{t-1}=2+(t-1)+\frac{2}{t-1} \\
\geqslant 2+2 \sqrt{(t-1) \frac{2}{t-1}}=2+2 \sqrt{2} .
\end{array}
$$

When $t-1=\frac{2}{t-1}$, i.e., $t=1+\sqrt{2}$, the equality holds.
Therefore, the maximum value of the real number $\lambda$ is $2+2 \sqrt{2}$.