Five. (20 points) Let two different tangent lines $l_{1}$ and $l_{2}$ be drawn from the point $P(1, -1)$ to the parabola $y = x^{2}$. Another tangent line to the parabola is $l$, and the angle (acute) between $l$ and $l_{1}$ is equal to the angle (acute) between $l$ and $l_{2}$. Try to find the equation of $l$.

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Let the tangent point of the tangent line $l_{1}$ or $l_{2}$ with the parabola $y=x^{2}$ be $\left(x_{0}, x_{0}^{2}\right)$. It is easy to know that the slope of the tangent line is $k=2 x_{0}$, and the equation of the tangent line is $y-x_{0}^{2}=2 x_{0}\left(x-x_{0}\right)$, i.e.,
$$
y=2 x_{0} x-x_{0}^{2} \text {. }
$$

Since $l_{1}$ and $l_{2}$ pass through the point $P(1,-1)$, from equation (1) we get
$$
-1=2 x_{0} \times 1-x_{0}^{2} \text {, }
$$

i.e., $\left(x_{0}-1\right)^{2}=2$.
Solving this, we get $x_{0}=1 \pm \sqrt{2}, k=2 \pm 2 \sqrt{2}$.
Let the tangent point of the tangent line $l$ with the parabola $y=x^{2}$ be $\left(x_{1}, x_{1}^{2}\right)$. It is easy to know that the slope of the tangent line $l$ is $2 x_{1}$, and the equation of $l$ is
$$
y-x_{1}^{2}=2 x_{1}\left(x-x_{1}\right),
$$

i.e., $y=2 x_{1} x-x_{1}^{2}$.
Since the angle (acute angle) between $l$ and $l_{1}$ is equal to the angle between $l$ and $l_{2}$, by the formula for the angle between two lines, we get
$$
\left|\frac{2 x_{1}-(2+2 \sqrt{2})}{1+2 x_{1}(2+2 \sqrt{2})}\right|=\left|\frac{2 x_{1}-(2-2 \sqrt{2})}{1+2 x_{1}(2-2 \sqrt{2})}\right| \text {, }
$$

i.e., $\left|\frac{x_{1}-(1+\sqrt{2})}{1+4(1+\sqrt{2}) x_{1}}\right|=\left|\frac{x_{1}-(1-\sqrt{2})}{1+4(1-\sqrt{2}) x_{1}}\right|$.
$$
\begin{array}{l}
\text { Hence }\left(x_{1}-1-\sqrt{2}\right)\left[1+4(1-\sqrt{2}) x_{1}\right] \\
=\left(x_{1}-1+\sqrt{2}\right)\left[1+4(1+\sqrt{2}) x_{1}\right],
\end{array}
$$

or $\left(x_{1}-1-\sqrt{2}\right)\left[1+4(1-\sqrt{2}) x_{1}\right]$
$$
=-\left(x_{1}-1+\sqrt{2}\right)\left[1+4(1+\sqrt{2}) x_{1}\right] \text {. }
$$

Simplifying, we get $4 x_{1}^{2}+1=0$ or $4 x_{1}^{2}+5 x_{1}-1=0$.
Therefore, only $x_{1}=\frac{-5 \pm \sqrt{41}}{8}, x_{1}^{2}=\frac{33 \mp 5 \sqrt{41}}{32}$.
Thus, the equation of $l$ is
$$
y=2 x_{1} x-x_{1}^{2}=\frac{-5 \pm \sqrt{41}}{4} \cdot x-\frac{33 \mp 5 \sqrt{41}}{32} .
$$