11. (15 points) Given real numbers $x, y \in (1, +\infty)$, and $xy - 2x - y + 1 = 0$. Find the minimum value of $\frac{3}{2} x^{2} + y^{2}$.
II. 11. From the problem, we know that $y=\frac{1}{x-1}+2$.
Then $\frac{3}{2} x^{2}+y^{2}=\frac{3}{2} x^{2}+\left(\frac{1}{x-1}+2\right)^{2}=f(x)$.
Thus, $f^{\prime}(x)=3 x+2\left(\frac{1}{x-1}+2\right)\left(-\frac{1}{(x-1)^{2}}\right)$
$=3 x-\frac{4 x-2}{(x-1)^{3}}$
$=\frac{(x-2)\left(3 x^{3}-3 x^{2}+3 x-1\right)}{(x-1)^{3}}$.
Let $g(x)=3 x^{3}-3 x^{2}+3 x-1$. Then
$g^{\prime}(x)=9 x^{2}-6 x+3>0$.
Also, $g(1)=6$, so $g(x)>0$ in the interval $(1,+\infty)$.
Therefore, $y=f(x)$ is monotonically decreasing in the interval $(1,2)$ and monotonically increasing in the interval $(2,+\infty)$.
Hence, $f(x)_{\min }=f(2)=15$.