Example 6 Find the equation of the curve $E^{\prime}$ symmetric to the curve $E: x^{2}+2 x y+y^{2}+3 x+y=$ 0 with respect to the line $l: 2 x-y-1=0$.
Solution: Let $M^{\prime}(x, y)$ be any point on the curve $E^{\prime}$, and the point symmetric to $M^{\prime}(x, y)$ with respect to the line $l$ be $M\left(x_{1}, y_{1}\right)$. By formula $(*)$ we get
$$
x_{1}=\frac{-3 x+4 y+4}{5}, y_{1}=\frac{3 y+4 x-2}{5} \text {, }
$$

Substituting into the equation of $E$ we get
$$
\begin{array}{l}
\left(\frac{-3 x+4 y+4}{5}+\frac{3 y+4 x-2}{5}\right)^{2} \\
+3\left(\frac{-3 x+4 y+4}{5}\right)+\frac{3 y+4 x-2}{5}=0
\end{array}
$$

Simplifying, we get $x^{2}+14 x y+49 y^{2}-21 x+103 y+54=0$, which is the desired equation.