Draw a tangent line to the parabola $y=x^{2}$ at point $A(1,1)$, which intersects the $x$-axis at $D$ and the $y$-axis at $B$. Point $C$ is on the parabola, and point $E$ is on the line segment $A C$, satisfying $\frac{|A E|}{|E C|}=\lambda_{1}$; point $F$ is on the line segment $B C$, satisfying $\frac{|B F|}{|F C|}=\lambda_{2}$, and $\lambda_{1}+\lambda_{2}=1$. The line segment $C D$ intersects $E F$ at point $P$. Find the equation of the trajectory of point $P$ as point $C$ moves along the parabola.
The slope of the tangent line passing through point $A$ on the parabola is: $y^{\prime}=\left.2 x\right|_{x=1}=2$, so the equation of the tangent line $A B$ is $y-2 x-1$. Therefore, the coordinates of $B$ and $D$ are $B(0,-1), D\left(\frac{1}{2}, 0\right)$, which means $D$ is the midpoint of segment $A B$.

Let $P(x, y), C\left(x_{0}, x_{0}^{2}\right), E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$. From $\frac{|A E|}{|E C|}=\lambda_{1}$, we have $x_{1}=$ $\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}, y_{1}-\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}} ; \frac{|B F|}{|F C|}=\lambda_{2}$, we get $x_{2}=\frac{\lambda_{2} x_{0}}{1+\lambda_{2}}, y_{2}=\frac{-1+\lambda_{2} x_{0}^{2}}{1+\lambda_{2}}$. Therefore, the equation of line $E F$ is:
$$
\frac{y-\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}}}{\frac{-1+\lambda_{2} x_{0}^{2}}{1+\lambda_{2}}-\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}}}=\frac{x-\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}}{\frac{\lambda_{2} x_{0}}{1+\lambda_{2}}-\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}},
$$

Simplifying, we get $\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}-\left(1+\lambda_{2}\right)\right] y-$
$$
\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}^{2}-3\right] x+1+x_{0}-\lambda_{2} x_{0}^{2} \text {. }
$$

When $x_{0} \neq \frac{1}{2}$, the equation of line $C D$ is: $y=$ $\frac{2 x_{0}^{2} x-x_{0}^{2}}{2 x_{0}-1}$ (2). Solving (1) and (2) together, we get $\left\{\begin{array}{l}x=\frac{x_{0}+1}{3} \\ y=\frac{x_{0}^{2}}{3}\end{array}\right.$, eliminating $x_{0}$, we obtain the trajectory equation of point $P$ as: $y=$ $\frac{1}{3}(3 x-1)^{2}$.
When $x_{0}=\frac{1}{2}$, the equation of $E F$ is: $-\frac{3}{2} y=\left(\frac{1}{4} \lambda_{2}-\frac{1}{4} \lambda_{1}-3\right) x+\frac{3}{2}-\frac{1}{4} \lambda_{2},$ and the equation of $C D$ is: $x=\frac{1}{2}$. Solving these together, we get $\left\{\begin{array}{l}x=\frac{1}{2}, \\ y=\frac{1}{12} .\end{array}\right.$, which is also on the trajectory of point $P$. Since $C$ cannot coincide with $A$, then $x_{0} \neq$ 1, hence $x \neq \frac{2}{3}$.
Therefore, the required trajectory equation is $y=\frac{1}{3}(3 x-1)^{2}\left(x \neq \frac{2}{3}\right)$.
From the first analysis, the equation of $A B$ is $y=2 x-1, \bar{B}(0,-1), D\left(\frac{1}{2}, 0\right)$, so $D$ is the midpoint of $A B$.

Let $\gamma=\frac{|C D|}{|C P|}, t_{1}=\frac{|C A|}{|C E|}=1+\lambda_{1}, t_{2}=\frac{|C B|}{|C F|}=1+\lambda_{2}$, then $t_{1}+t_{2}=3$. Since $C D$ is the median of $\triangle A B C$, we have $S_{\triangle C A B}=2 S_{\triangle C A D}=2 S_{\triangle C B D}$. And
$$
\begin{aligned}
\frac{1}{t_{1} t_{2}} & =\frac{|C E| \cdot|C F|}{|C A| \cdot|C B|}=\frac{S_{\triangle C E F}}{S_{\triangle C A B}}=\frac{S_{\triangle C E P}}{2 S_{\triangle C A D}}+\frac{S_{\triangle C F P}}{2 S_{\triangle C B D}} \\
& =\frac{1}{2}\left(\frac{1}{t_{1} \gamma}+\frac{1}{t_{2} \gamma}\right)=\frac{t_{1}+t_{2}}{2 t_{1} t_{2} \gamma}=\frac{3}{2 t_{1} t_{2} \gamma},
\end{aligned}
$$

Then $\gamma=\frac{3}{2}$, so $P$ is the centroid of $\triangle A B C$.
Let $P(x, y), C\left(x_{0}, x_{0}^{2}\right)$. Since point $C$ is different from $A$, then $x_{0} \neq 1$, hence the coordinates of the centroid $P$ are $x=$ $\frac{0+1+x_{0}}{3}=\frac{1+x_{0}}{3}\left(x \neq \frac{2}{3}\right), y=\frac{-1+1+x_{0}^{2}}{3}=\frac{x_{0}^{2}}{3}$, eliminating $x_{0}$, we get $y=$ $\frac{1}{3}(3 x-1)^{2}$. Therefore, the required trajectory equation is $y=\frac{1}{3}(3 x-1)^{2}\left(x \neq \frac{2}{3}\right)$.