find the min value

Let $u=x-1$ and $v=y-1$ . Note that $u,v >0$ , so Titu's lemma and AM-GM inequality applies.

$\begin{aligned} \frac {x^2}{y-1} + \frac {y^2}{x-1} & = \blue{\frac {(u+1)^2}v + \frac {(v+1)^2}u} & \small \blue{\text{By Titu's lemma}} \\ \blue{\frac {(u+1)^2}v + \frac {(v+1)^2}u} & \ge \red{\frac {(u+v+2)^2}{u+v}} & \small \red{\text{Let }z=u+v} \\ \red{\frac {z^2 + 4z+4}z} & = z + 4 + \frac 4z \\ \blue{z + \frac 4z} + 4 & \ge \blue{2\sqrt 4} + 4 = \boxed 8 & \small \blue{\text{By AM-GM inequality}} \end{aligned}$

Equality occurs when $x=y=2$ .

The symmetry of the expression indicates that the expression will attain an optimum when $x=y$ . Then, the optimum value of the expression is $\dfrac{2x^2}{x-1}=c$ (say). This gives rise to the equation

$2x^2-cx+c=0$ . Since $x$ is real, therefore $c^2\geq 4\times 2\times c\implies c\geq 8$ . Hence the minimum value of the expression is $\boxed 8$ .

Alternative method :

Let $x=a+1, y=b+1$ . Then the given expression is $\dfrac{a^2+2a+1}{b}+\dfrac{b^2+2b+1}{a}=(\frac{a^2}{b}+\frac{b^2}{a})+2(\frac{a}{b}+\frac{b}{a})+(\frac{1}{a}+\frac{1}{b})$ . By A. M. - G. M. inequality, $\frac{a^2}{b}+\frac{b^2}{a}\geq 2\sqrt {ab}, \frac{a}{b}+\frac{b}{a}\geq 2, \frac{1}{a}+\frac{1}{b}\geq \frac{2}{\sqrt {ab}}, \sqrt {ab}+\frac{1}{\sqrt {ab}}\geq 2$ . Hence the given expression is $\geq 2\times 2+2\times 2=8$ . Therefore the minimum value of the given expression is $\boxed 8$ .