Minimum too?

Call the expression P, multiply everything out and we get $P=16(xy)^2+12(x^3+y^3)+9xy+25xy=16(xy)^2+12(x+y)^2-36xy+34xy$ $\Leftrightarrow P=16(xy)^2-2xy+12$ Now we let $t=xy$ , then $f(t)=16t^2-2t+12$ From the condition, we can prove that $t\in \big[0;\frac{1}{4}\big]$ , we see that $P=16\bigg(t-\frac{1}{16}\bigg)^2+\frac{191}{16}\geq\frac{191}{16}$ The equality holds when $t=\frac{1}{16}$ which is in the range of $t$

Now the only thing we have to do left is compare $f(0)$ with $f\big(\frac{1}{4}\big)$ . $f\big(\frac{1}{4}\big)$ is larger so $P\leq f\bigg(\frac{1}{4}\bigg)=\frac{25}{2}$ $\therefore\frac{m}{n}=\frac{191}{16}+\frac{25}{2}=\frac{391}{16} \Rightarrow m+n=407$ P reaches its minimum when $(x,y)=\big(\frac{2-\sqrt{3}}{4};\frac{2+\sqrt{3}}{4}\big)$ and its permutation, reaching its maximum when $x=y=\frac{1}{2}$

$\begin{aligned} X & = (4x^2+3y)(4y^2+3x) + 25xy \\ & = 16x^2y^2 + 12x^3 + 12y^3 + 9xy + 25xy \\ & = 16x^2y^2 + 12(x^3 + y^3) + 34xy \\ & = 16x^2y^2 + 12(x+y)(x^2 + y^2-xy) + 34xy \\ & = 16x^2y^2 + 12(x^3 + y^3) + 34xy \\ & = 16x^2y^2 + 12(x+y)((x+y)^2-2xy-xy) + 34xy \\ & = 16x^2y^2 + 12(1-3xy) + 34xy \\ & = 16x^2y^2 -2xy + 12 \\ \implies X & = \left(4xy - \frac{1}{4}\right)^2 + \frac{191}{16} \end{aligned}$

Since $\left(4xy - \dfrac{1}{4}\right)^2 \ge 0$ , $X_{min} = \dfrac{191}{16}$ .

$X$ is maximum when $4xy - \dfrac{1}{4}$ is maximum. By AM-GM inequility, $x+y \ge 2\sqrt{xy}$ $\implies 2\sqrt{xy} \le 1$ $\implies xy \le \dfrac{1}{4}$ . Therefore, $X_{max} = \left(4\times \dfrac{1}{4} - \dfrac{1}{4}\right)^2 + \dfrac{191}{16} = \dfrac{200}{16}$ .

And that $X_{max} + X_{min} = \dfrac{391}{16}$ $\implies m + n = \boxed{407}$