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 $xy$ plane. Find its center of symmetry and axes of symmetry.

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Three, it is known that points $A(1,1), B(1,-1)$ are on the curve.
Let the center of symmetry of the ellipse be $(a, b)$, then points $A^{\prime}(2 a-1, 2 b-1), B^{\prime}(2 a-1,2 b+1)$ are also on the curve. Therefore, we have
$$
\begin{array}{l}
17(2 a-1)^{2}-16(2 a-1)(2 b-1)+4(2 b-1)^{2}- \\
34(2 a-1)+16(2 b-1)+13=0, \\
17(2 a-1)^{2}-16(2 a-1)(2 b+1)+4(2 b+1)^{2}- \\
34(2 a-1)+16(2 b+1)+13=0 .
\end{array}
$$
Subtracting (2) from (1) and simplifying gives $b=2 a-2$. Substituting $b=2 a-2$ into (1) and simplifying yields
$$
4 a^{2}-8 a+4=0 \text {. }
$$

Solving for $a$ gives $a=1$. Thus, $b=0$.
Hence, the center of symmetry is $(1,0)$.
Since the axis of symmetry passes through the center of symmetry, we can assume the equation of the axis of symmetry is $y=k(x-1)$. Let the point $A(1,1)$ be symmetric to point $A_{0}\left(x_{0}, y_{0}\right)$ with respect to the line $y=k(x-1)$, then we have
$$
\left\{\begin{array} { l } 
{ \frac { y _ { 0 } + 1 } { 2 } = k ( \frac { x _ { 0 } + 1 } { 2 } - 1 ) , } \\
{ \frac { y _ { 0 } - 1 } { x _ { 0 } - 1 } = - \frac { 1 } { k } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{0}=\frac{1+2 k+k^{2}}{1+k^{2}}, \\
y_{0}=\frac{k^{2}-1}{1+k^{2}} .
\end{array}\right.\right.
$$

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

Substituting (3) into the above equation and simplifying gives
$$
k\left(8 k^{2}-13 k-8\right)=0 \text {. }
$$
$k=0$ is not a valid solution, hence
$$
\begin{array}{l}
8 k^{2}-13 k-8=0 \\
\Rightarrow k=\frac{13 \pm 5 \sqrt{17}}{16} .
\end{array}
$$

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