3.4.1. Find the equations of the tangents to the curve $y=\frac{3 x+2}{3 x-2}$ that are parallel to the line $3 x+y+3=0$.
Solution. Let $M\left(x_{0} ; y_{0}\right)$ be the coordinates of the point of tangency of the tangent and the curve. We choose the equation of the tangent in the form:

$$
y-y_{0}=k_{k a c .} \cdot\left(x-x_{0}\right)
$$

Since the tangent is parallel to the line $3 x+y+3=0$, the slope of which is $k_{n p .}=-3$, then $k_{k a c .}=k_{n p .}=-3$.

Let's find the derivative:

$$
\begin{gathered}
y^{\prime}=\frac{3 x+2}{3 x-2}=\frac{(3 x+2)^{\prime}(3 x-2)-(3 x+2)(3 x-2)^{\prime}}{(3 x-2)^{2}}= \\
=\frac{3(3 x-2-3 x-2)}{(3 \cdot x-2)^{2}}=\frac{-12}{(3 x-2)^{2}} .
\end{gathered}
$$

Since the derivative $y^{\prime}$ at the point of tangency $\left(x_{0} ; y_{0}\right)$ is numerically equal to the slope of the tangent, then

$$
y^{\prime}\left(x_{0}\right)=\frac{-12}{\left(3 x_{0}-2\right)^{2}}=K_{k a c .}=-3 \text {, }
$$

from which

$$
-12=-3\left(3 x_{0}-2\right)^{2} \text { and } 9 x_{0}^{2}-12 x_{0}=0 \text {. }
$$

We get

$$
x_{01}=0, x_{02}=\frac{4}{3}
$$

then

$$
y_{01}=\frac{3 \cdot 0+2}{3 \cdot 0-2}=-1 \text { and } y_{02}=\frac{3 \cdot \frac{4}{3}+2}{3 \cdot \frac{4}{3}-2}=3
$$

We have two points of tangency $M_{1}(0 ;-1)$ and $M_{2} \frac{4}{3} ; 3$, accordingly to which we will write two equations of tangents: $y+1=-3(x-0)$, from which

$$
3 x+y+1=0 \text { and } y-3=-3 x-\frac{4}{3},
$$

from which

$$
3 x+y-7=0 .
$$