Example 1 Let $m$ be a real number not less than -1, such that the equation in $x$
$$
x^{2}+2(m-2) x+m^{2}-3 m+3=0
$$

has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$, find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$.
(2000, National Junior High School Mathematics Competition)
Explanation: Since the equation has two distinct real roots, we have
$$
\begin{array}{l}
\Delta=4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
=-4 m+4>0 .
\end{array}
$$

Thus, $m<1$.
Also, $m \geqslant-1$, so, $-1 \leqslant m<1$.
By the relationship between roots and coefficients, we get
$$
x_{1}+x_{2}=-2(m-2), x_{1} x_{2}=m^{2}-3 m+3 \text {. }
$$
(1) Since $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$
$$
=2 m^{2}-10 m+10=6 \text {, }
$$

Then we have $2 m^{2}-10 m+4=0$.
Solving the equation and considering $-1 \leqslant m<1$, we get
$$
m=\frac{5-\sqrt{17}}{2} \text {. }
$$
$$
\begin{array}{l}
\text { (2) } \frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}} \\
=\frac{m x_{1}^{2}\left(1-x_{2}\right)+m x_{2}^{2}\left(1-x_{1}\right)}{\left(1-x_{1}\right)\left(1-x_{2}\right)} \\
=\frac{m\left(x_{1}^{2}+x_{2}^{2}\right)-m x_{1} x_{2}\left(x_{1}+x_{2}\right)}{1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}} \\
=2\left(m^{2}-3 m+1\right) . \\
\text { Let } y=2\left(m^{2}-3 m+1\right) \\
=2\left(m-\frac{3}{2}\right)^{2}-\frac{5}{2}(-1 \leqslant m<1) .
\end{array}
$$

According to the graph of the function $y$, $y$ is monotonically decreasing on $-1 \leqslant m<1$.

Thus, when $m=-1$, $y$ reaches its maximum value of 10, i.e., the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$ is 10.