Find $h(x)$

Since $\begin{aligned} \ & F(x) = \frac{x^2+1}{x^2} \\& \implies F(h(x)) = \frac{(h(x))^2+2}{(h(x))^2} \\&\implies x(h(x))^2 - (h(x))^2 =2 \\ & \implies h(x) = \sqrt{\frac{2}{x-1}} \\& \implies h(3) = \pm 1\end{aligned}$

To make things simpler, let $y=h(x)$ . Since $F(h(x))=x$ , we set up an equation by substituting $h(x)~(\text{or}~y)$ for $x$ , and setting the whole thing equal to $x$ : $\frac{y^2+2}{y^2}=x$ So we need to solve for $y$ to get the function. Then, we can plug 3 in. In order for the first step to occur, we should simplify the fraction: $1+\frac{2}{y^2}=x$ Then, we subtract 1 from both sides and multiply by $y^2$ : $2=y^2(x-1)$ The steps are quite obvious from here: $\frac{2}{x-1}=y^2$ $y=\pm\sqrt{\frac{2}{x-1}}$ Then we plug in 3 for $x$ and $h(3)$ for $y$ : $h(3)=\pm\sqrt{\frac{2}{3-1}}=\pm\sqrt{1}=\boxed{\pm1}$

$\begin{aligned} F(h(x)) & = x & \small \color{#3D99F6} \text{Given} \\ \implies F(h(3)) & = 3 \\ \implies \frac {(h(3))^2+2}{(h(3))^2} & = 3 \\ (h(3))^2 + 2 & = 3(h(3))^2 \\ 2(h(3))^2 & = 2 \\ (h(3))^2 & = 1 \\ \implies h(3) & = \boxed{\pm 1} \end{aligned}$

$F(h(x)) = x \implies h(x) \ \ \text {is the inverse of } \ \ F(x)$

$\text{Given that} \ \ F(x) = \dfrac{x^2 +2} {x^2} = Y \ \ \ \text {then} \ \ x =\pm \sqrt{\dfrac {2}{Y-1}}$

$\text{Or} \ \ h(x) = \pm \sqrt{\dfrac {2}{x-1}} \ \ \text{ and} \ \ h(3) = \pm 1$