Three. (20 points) Given the equation
$$
17 x^{2}-16 x y+4 y^{2}-34 x+16 y+13=0
$$

represents an ellipse in the $x O y$ plane. Find its center of symmetry and axes of symmetry.
Three, it is known that points $A(1,1), B(1,-1)$ are on the ellipse.
Let the center of symmetry of the ellipse be $(a, b)$.
Then, by symmetry, the points
$$
A^{\prime}(2 a-1,2 b-1), B^{\prime}(2 a-1,2 b+1)
$$

are also on the ellipse.
Substituting into the known equation, we get
$$
\begin{array}{l}
17(2 a-1)^{2}-16(2 a-1)(2 b-1)+ \\
4(2 b-1)^{2}-34(2 a-1)+16(2 b-1)+13 \\
=0, \\
17(2 a-1)^{2}-16(2 a-1)(2 b+1)+ \\
4(2 b+1)^{2}-34(2 a-1)+16(2 b+1)+13 \\
=0 .
\end{array}
$$
Subtracting (2) from (1) and simplifying, we get
$$
b=2 a-2 \text {. }
$$

Substituting equation (3) into equation (1) and simplifying, we get
$$
4 a^{2}-8 a+4=0 \Rightarrow a=1, b=0 \text {. }
$$

Thus, the center of symmetry of the ellipse is $(1,0)$.
Since the axis of symmetry passes through the center of symmetry, we can set the equation of the axis of symmetry as
$$
y=k(x-1)(k \neq 0) .
$$

Let the point $A(1,1)$ be symmetric to the point $A_{0}\left(x_{0}, y_{0}\right)$ with respect to the line $y=k(x-1)$. Then
$$
\begin{aligned}
& \left\{\begin{array}{l}
\frac{y_{0}+1}{2}=k\left(\frac{x_{0}+1}{2}-1\right), \\
\frac{y_{0}-1}{x_{0}-1}=-\frac{1}{k}
\end{array}\right. \\
\Rightarrow & \left\{\begin{array}{l}
x_{0}=\frac{k^{2}+2 k+1}{k^{2}+1}, \\
y_{0}=\frac{k^{2}-1}{k^{2}+1} .
\end{array}\right.
\end{aligned}
$$

Since point $A_{0}$ is on the ellipse, then
$$
17 x_{0}^{2}-16 x_{0} y_{0}+4 y_{0}^{2}-34 x_{0}+16 y_{0}+13=0 \text {. }
$$

Substituting equation (4) into the above equation and simplifying, we get
$$
k\left(8 k^{2}-13 k-8\right)=0 \text {. }
$$

Since $k \neq 0$, we have
$$
8 k^{2}-13 k-8=0 \Rightarrow k=\frac{13 \pm 5 \sqrt{17}}{16} \text {. }
$$

Therefore, the equation of the axis of symmetry of the ellipse is
$$
y=\frac{13 \pm 5 \sqrt{17}}{16}(x-1) .
$$