What's There to Rationalize?

Since $x = \sqrt{3}+\sqrt{2}$ and $y = \sqrt{3}-\sqrt{2}$ , $x$ and $y$ are the roots of $w^2 - 2\sqrt{3}w + 1 = 0$ . Because $w = \dfrac{2\sqrt{3} \pm \sqrt{12-4}}{2} = \sqrt{3} \pm \sqrt{2}$ .

This implies that $\begin{cases} x^2 - 2\sqrt{3}x + 1 = 0 & \Rightarrow x^2 + 1 = 2\sqrt{3}x \\ y^2 - 2\sqrt{3}y + 1 = 0 & \Rightarrow y^2 + 1 = 2\sqrt{3}y \\ x + y = 2\sqrt{3} \\ xy = 1 \end{cases}$

Therefore, we have:

$\begin{aligned} \frac{y}{x^2+1} + \frac{x}{y^2+1} & = \frac{y}{2\sqrt{3}x} + \frac{x}{2\sqrt{3}y} \\ & = \frac{1}{2\sqrt{3}} \left( \frac{y}{x} + \frac{x}{y} \right) \\ & = \frac{1}{2\sqrt{3}} \left( \frac{y^2+x^2}{xy} \right) \\ & = \frac{1}{2\sqrt{3}} \left( \frac{(x+y)^2 -2xy}{1} \right) \\ & = \frac{1}{2\sqrt{3}} \left((2\sqrt{3})^2 -2(1) \right) \\ & = \frac{10}{2\sqrt{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \end{aligned}$

$\Rightarrow a+b = 5+3 = \boxed{8}$